Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases.
 A: $$
e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n\ge1+x
$$
by Bernoulli's inequality.
A: We want to prove that $1+x\le e^x$ for any $x\in\mathbb R$.  Setting $x=\log(u)$, this is equivalent to proving: 
$$
1+\log(u)\le u
$$
for any $u\in (0, \infty)$.  
This is true because: 
$$
1+\log(u)=1+\int_1^u\frac1tdt\le1+\int_1^u1dt=1+u-1=u
$$
Some care is needed to establish that the inequality is true for both $u\ge1$ and $0<u\le1$.  In the second case, we can see this more clearly by writing: 
$$
1+\int_1^u\frac1tdt=1+\int_u^1-\frac1tdt\le1+\int_u^1-1dt=1-1+u=u
$$
A: The shortest proof I could think of:
$$1 + x \leq 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = e^x.$$
However, it is not completely obvious for negative $x$.
Using derivatives:
Take $f(x) = e^x - 1 - x$. Then $f'(x) = e^x - 1$ with $f'(x) = 0$ if and only if $x = 0$. But this is a minimum (global in this case) since $f''(0) = 1 > 0$ (the second derivative test). So $f(x) \geq 0$ for all real $x$, and the result follows.
Another fairly simple proof (but it uses Newton's generalization of the Binomial Theorem which is often covered in precalculus):
We proceed by contradiction. Suppose the inequality does not hold, i.e., $e^x < 1 + x$ for some $x$. Then $e^{kx} < (1 + x)^k$. Now set $x = 1/k$ so that
\begin{align*}
e &< \left( 1 + \frac{1}{k} \right)^k\\
&= 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k(k - 1)}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k(k - 1)(k - 2)}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\
&< 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k^2}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k^3}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\
&= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\\
&= e,
\end{align*}
which is absurd. Therefore $1 + x \leq e^x$ for all real $x$.
By the way, this is where
$$e = \lim_{k \to \infty}\left( 1 + \frac{1}{k} \right)^k$$
comes from because
$$\lim_{k \to \infty}\frac{k(k - 1)}{k^2} = \lim_{k \to \infty}\frac{k(k - 1)(k - 2)}{k^3} = \cdots = \lim_{k \to \infty}\frac{k(k - 1)(k - 2) \cdots (k - n)}{k^{n + 1}} = \cdots = 1.$$
A: There is an amusing proof that I found yesterday that $e^x>x$ for every $x\in \mathbb{R}$.
It is obvious that $e^x>x$ if $x<0$ since the LHS is positive and the RHS is negative.
Suppose that for some $a\ge 0$, the inequality $e^a\le a$ holds.
Then $a\ge e^a\ge 1$ since $e^a\ge e^0$ because $a\ge 0$. But now we can see that $a\ge 1$ and again, $a\ge e^a\ge e^1$ and so $a\ge e$. We continue applying the same observation and conclude that $a\ge e^{^{e}}$ and so on, which means that $a$ is unbounded which is a contradiction.
A: For positive values ​​of $ x $  We can use the following characterization of $e^x$
$$
e^x=\lim_{t\to \infty} \Big( 1+\frac{1}{t}\Big)^{tx},\quad t> 0,x\geq 0.
$$
The Bernoulli's inequality states that $(1 + y)^r \geq 1 + ry$ for every $r \geq 1$ and every real number $y \geq −1$. Then for $y=\frac{1}{t}$ and $t>0$ such that $r=tx\geq 1$ we have
\begin{align}
e^x= &\lim_{t\to \infty} \Big( 1+\frac{1}{t}\Big)^{tx}\\
\geq &\lim_{t\to \infty} \Big(1+\frac{1}{t}(tx) \Big)\\
 = & 1+x
\end{align}
A: Let $f(x) = e^x-(1+x)$, then $f^\prime(x) = e^x-1$. Hence $f^\prime(x)=0$ iff $x=0$. Furthermore $f^{\prime\prime}(0) = e^0=1>0$, thus $f(0)=0$ must be the global minimum of $f$, proving your claim.
A: One which uses $\exp(x) = \frac 1{\exp(-x)} $
$$   1 + x \underset{ \text{obvious}\\ \text{for $x>0$}}{\lt}  1 + x + {x^2 \over 2!} + {x^3 \over 3!} + ... = {1 \over   1  - x + {x^2 \over  2!} - { x^31 \over 3!} + ...  } \tag 1 $$
Now we replace $+x$ by its negative counterparts and get similarily
$$   1 - x \underset{ \quad \text{for $x>0$}\\ \text{but not obvious}}{\lt}  1 - x + {x^2 \over 2!} - {x^3 \over 3!} + ... = {1 \over   1  + x + {x^2 \over  2!} + {x^3 \over  3!} + ...  }\tag 2$$
But now the comparision with the fraction on the rhs becomes obvious if we look at the reciprocals. The reciprocal  ${1\over 1-x}=1+x+x^2+x^3+...$ is and we get
$$   {1 \over 1 - x} = 1+x+x^2+... \underset{ \text{obvious}\\ \text{for $x>0$}}{\gt} 
      1  + x + {x^2 \over  2!} + {x^3 \over  3!} + ... = {1 \over 1 - x + {x^2 \over 2!} - {x^3 \over 3!} + ... }\\ \tag 3 $$
A: One that's not been mentioned so far(?): knowing that
$$
0 < e^x = 
 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots
$$
proves the inequality except for $-1 < x < 0$.  But in that region
$$
e^x - (1+x) = 
 \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots
$$
is an alternating series whose terms decrease in absolute value
and start out positive.  Therefore it is positive by the usual argument:
group the terms as
$$
e^x - (1+x) =
\left( \frac{x^2}{2} + \frac{x^3}{3!} \right) + 
\left( \frac{x^4}{4!} + \frac{x^5}{5!} \right) + \cdots
$$
and observe that each combined term is positive, QED.
(This actually works for $-3 < x < 0$, but you still want to use
$e^x > 0$ to prove the inequality for very negative $x$.)
A: Proof by induction (works for natural numbers)
Assume it works for n 
1 + n < e^n

Then we prove that it works  for n+1 
1 + (n+1) < e^(n+1)  

Proof
         1 + n < e^n 
  or    1 + n + 1 < e^n + 1 
  or    1 + n + 1 < e^n + e^n   since e^n > 1
  or    1 + n + 1 < e^n * 2
  or    1 + (n+1) < e^n * e       since e > 2
  or    1 + (n+1) < e^(n+1)

hence it is true for n+1 if true for n. We know it is true for 1, hence by induction is true of 2, 3, 4...so on.
A: For $x>0$ we have  $e^t>1$ for $0<t<x$
Hence, $$x=\int_0^x1dt \color{red}{\le}  \int_0^xe^tdt =e^x-1 \implies  1+x\le e^x$$
For $x<0$ we have  $e^{t} <1$ for $x <t<0$
$$-x=\int^0_x1dt \color{red}{\ge} \int^0_xe^tdt =1-e^x \implies  1+x\le e^x$$
A: Another simple proof...
Define function $f(x)=e^x-(x+1)$. The minimum value is $0$ at $x=0$, it's also convex, so $f(x) \ge 0$.
A: Repeatedly using $1 + x \le \left(1 + \frac{x}{2} \right)^2$, we have
\begin{align}
1 + x
\le
\left(1 + \frac x 2\right)^2
\le
\left(1 + \frac x 4\right)^4
\le
\left(1 + \frac x 8\right)^8
\le
\dots
\le
\left(1 + \frac x {2^k}\right)^{2^k}.
\end{align}
Taking the limit of $k \rightarrow \infty$ yields
$$
1 + x \le e^x. \qquad\qquad(1)
$$

Another proof using the technique in this post.  By the AM-GM inequality,
$$
\sqrt[n]{1 \times \cdots \times 1 \times (1 + x)}
\le
\frac{1 + \dots + 1 + (1 + x)}{n}
=1 + \frac{x}{n}.
$$
So,
$$
1+x \le \left(1 + \frac{x}{n} \right)^n.
$$
Taking the limit of $n \rightarrow \infty$ yields (1).
A: If $x \ge 0$ then
$$\begin{align}
e^x = 1 + \int_0^x e^t\,\mathrm dt &= 1 + \int_0^x\left( 1 + \int_0^t e^u \,\mathrm du\right)\,\mathrm dt \\&= 1+x + \int_0^x \int_0^t e^u\,\mathrm du\,\mathrm dt \ge 1+x\end{align}$$
If $x \le 0$ then $$\begin{align}e^x = 1 - \int_x^0 e^t\,\mathrm dt &= 1 - \int_x^0\left( 1 - \int_t^0 e^u\,\mathrm du\right)\,\mathrm dt \\ &= 1+x + \int_x^0 \int_t^0 e^u\,\mathrm du\,\mathrm dt \ge 1+x\end{align}$$
A: For $x > 0$, consider the mean value theorem on the interval $[0,x]$. Then
$$e^x - e^0 \geq \inf_{(0,x)} e^c \cdot (x - 0) = x,$$
implying $e^x \geq 1+x$. For $x < 0$, apply MTV on $[x,0]$:
$$e^0 - e^x \leq \sup_{(x,0)} e^c \cdot (0 - x) = -x,$$
giving us $-e^x \leq -x - 1$, or $e^x \geq x + 1$. Then check $x = 0$ and the equality is proven.
A: Another way (not sure if its "simple" though!):  $y = x+1$ is the tangent line to $y = e^x$ when $x= 0$.  Since $e^x$ is convex, it always remains above its tangent lines.
A: For completeness, using $\exp(x)=1+x+\frac{1}{2}x^2+\dots$, the inequality is trivial for $x\ge 0$. It is also trivial for $x<-1$. 
It remains to show the case $-1<x<0$. Replacing $x$ by $-x$, one need to show $1-x < e^{-x}$ for $0<x<1$, or 
$$1+x+\frac{1}{2}x^2+\dots=e^x <\frac{1}{1-x}=1+x+x^2+\dots,$$ 
we are done.
A: Completing glebovg's answer :


*

*the inequality $1+x \le e^x$ clearly holds for $x \leq -1$,

*suppose $x \geq -1$ :
the series $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ can be written (grouping the terms in pair) :
$$e^x = 1 + x + \sum_{k \geq 1} \left( \frac{x^{2k}}{(2k)!} + \frac{x^{2k+1}}{(2k+1)!} \right)$$
$$e^x = 1 + x + \sum_{k \geq 1} x^{2k}\left( \frac{1}{(2k)!} + \frac{x}{(2k+1)!} \right)$$
$$e^x = 1 + x + \sum_{k \geq 1} x^{2k}\left( \frac{2k + 1 + x}{(2k+1)!} \right)$$
under the assumption $x \geq -1$, the $\sum$ part is clearly a sum of positive numbers.
A: Beautiful answers, but nobody used The Mean Value Theorem.  Apply MVT on $[0,x] $ for $x>0$. There is some $c\in (0,x)$ such that:
\begin{align} 
\frac{e^x-e^0}{x-0} = e^c > 1
\end{align} 
So
\begin{align} 
e^x>1+x
\end{align} 
Something similar can be done for $x<0$. Finally note that we have equality when $x=0$. So we get the desired result:
\begin{align} 
e^x\geq 1+x 
\end{align} 
A: Let $f(x)=\exp(x)-x-1$. Then, $f'(0)=0$. But $f$ is strictly convex (a difference of a strictly convex function and an affine one), so that $0$ most be a unique global minimum. Hence, $\exp(x)-x-1=f(x)\geq f(0)=0$ for all $x\in\mathbb{R}$.
A: We want to show that (1) $$1+x\leq e^x,$$ for $x\in\mathbb{R}$. When $x\geq 0$, we have $$1+x\leq 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=e^x.$$ Suppose $x=-X$, where $X>1$, then $1+x=1-X<0$ and $e^{x}=e^{-X}=1/e^X>0$. Hence (1) holds.
Now take logarithms of (1) to obtain $$\log(1+x)\leq x.$$ But $$\log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots,$$ where $|x|<1$. Suppose $x=-X$, where $0<X<1$, then $$\log(1+x)=\log(1-X)=-X-\frac{X^2}{2}-\frac{X^3}{3}-\cdots<-X=x.$$ Or, equivalently, $$1+x<e^x,$$ where $-1<x<0$.
A: The fact $\frac{d}{dx} e^x = e^x$ is nicely demonstrated using the self-similar nature of exponential functions. (See my answer here.)
This justifies (actually, declares) that $y=x+1$ is tangent to $y=e^x$; thereafter, since the slope increases (or decreases) as $x$ gets larger (respectively, smaller) the line and curve cannot meet again (which is an informal way of stating the convexity property).
A: A proof using only a little basic calculus, and not too many cases:
Set
$\alpha(x) = e^{-x}(1 + x); \tag{1}$
then
$\alpha'(x) = -e^{-x}(1 + x) + e^{-x} = -xe^{-x}, \tag{2}$
and
$\alpha(0) = 1; \tag{3}$
we note that
$x > 0 \Rightarrow \alpha'(x) < 0 \tag{4}$
and
$x < 0 \Rightarrow \alpha'(x) > 0 \tag{5}$
with
$\alpha'(0) = 0; \tag{6}$
thus, for $x > 0$,
$\alpha(x) - 1 = \alpha(x) - \alpha(0)  = \int_0^x \alpha'(s) ds < 0, \tag{7}$
whence
$e^{-x}(1 + x) = \alpha(x) < 1; \tag{8}$
when $x < 0$,
$1 - \alpha(x) = \int_x^0 \alpha'(s) ds > 0, \tag{9}$
yielding
$e^{-x}(1 + x) = \alpha(x) < 1 \tag{10}$
in this case as well; (8) and (10) together imply 
$1 + x < e^x \tag{11}$
when $x \ne 0$; clearly
$1  + 0 = 1 = e^0; \tag{12}$
combining (11) and (12) shows that
$1 + x \le e^x \tag{13}$
for every $x \in \Bbb R$, with strict inequality precisely when $x \ne 0$.   QED.
A: We know the function $x^x$ has a single local minimum at $x=\frac1e$. Thus, for positive $x$, we have:
\begin{align}
\left(\frac1e\right)^{1/e}&\le x^x\\
e^{1/e}&\ge\frac1{x^x}\\
e^{1/xe}&\ge\frac1x\\
e^{1/xe-1}&\ge\frac1{xe}\\
e^{(1/xe-1)}&\ge\left(\frac1{xe}-1\right)+1
\end{align}
Let $t=\frac1{xe}-1$. If $x>0$, we have $t>-1$. Thus, for all $t>-1$:
$$e^t\ge t+1$$
(To prove the above for $t\le -1$, simply note that the left-hand side is always positive while the right-hand side would be zero or negative.) QED.
A: The approximation of the exponential function by its linear Taylor polynomial has remainder term $R(x) := \exp(x) - (1+x)$. The Taylor Remainder Theorem then yields some $\xi$ in between $0$ and $x$ such that $R(x) = \exp''(\xi) x^2 = \exp(\xi) x^2 \ge 0$.
A: The series expansion of $(1+x)$ is $(1+x)$, while $\exp(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+...$. Subtracting the second from the first one you have the difference $d=\exp(x)-(1+x)=\frac{1}{2}x^2+\frac{1}{6}x^3+...$ which is zero only for $x=0$ otherwise $d\gt 0$
Q.E.D.
A: This answer uses no calculus or geometry.
Prerequisites: Algebra, basic facts about limits and that for $a \gt 0$ we can define $a^x$ for $x \in \Bbb R$
(see limits of rational exponents).

Theorem 1: There is one and only one number $a \gt 0$ satisfying
$$\tag 1 \forall x \in \Bbb R, \; a^x \ge 1 + x$$

Analyzing $\text{(1)}$, you'll be naturally lead to examine
$\tag 2 u_n \le a \le v_n \text{ where } n \ge 2$
with
$\tag 3 u_n = (1 + \frac{1}{n})^n \text{ and } \le  v_n = (1 - \frac{1}{n})^{-n}$
Searching, you find answer links from this site:
$\quad u_n \text{ is strictly increasing}:\quad$ here
$\quad v_n \text{ is algebraically related to } u_n:\quad$ here
$\quad u_n \le 3:\quad$ here
By working with the theory in the above links you will conclude that only one real number, call it $e$, can possibly satisfy $\text{(1)}$. Again, as in the first answer link above, you will use the Bernoulli's inequality and
$\tag 4 e^\frac{s}{t} =\lim_{n\to \infty} \Big( 1+\frac{\frac{s}{t}}{\frac{ns}{t}} \Big)^{\frac{ns}{t}}$
to wrap things up:
$$ \forall x \in \Bbb R, \; e^x \ge 1 + x$$
A: $1+x \le e^x$ 
Take the ln of both sides
$\ln(1+x) \le x$ 
differentiate both sides w.r.t $x$
$\frac{1}{1+x} \le 1$
which holds $\forall x \in \mathbb{R}, x \neq -1$.
