Proof of $1+x\leq e^x$ for all x? Does anyone provide proof  of $1+x\leq e^x$ for all $x$?
What is the minimum $a(>0)$  such that $1+x\leq a^x$ for all x?
 A: The strict inequality should be $\le$, as pointed out in the comments.
One way is to look at the derivative of $f(x) = e^x - x - 1$, which is $f'(x) = e^x - 1$, and note that it is zero only at $x=0$. The second derivative is $f''(x)=e^x > 0$ for all $x$, so $x=0$ is a global minimizer. Finally, note $f(0) = 0$, which yields $e^x-x-1 = f(x) \ge 0$ for all $x$.
A: While other answers have tried to give different proofs of the inequality, I will deal with the more important and slightly difficult second part where it asks for minimum value of $a$ such that $a^{x} \geq 1+x$.
Well it turns out that there is one and only one value of $a$ such that $a^{x} \geq 1+x$ and that value is $a=e$. We have the following theorem:

Theorem: Let $f:\mathbb{R} \to\mathbb{R} $ be a function such that $f(x) \geq 1+x$ for all $x\in \mathbb {R} $ and further $f(x+y) =f(x) f(y) $ for all $x, y\in\mathbb {R} $. Then $f(x) =\exp(x)$.

Proof: First some obvious observations. From the inequality $f(x) \geq 1+x$ we can see that $f(x) >0$ for all $x\geq 0$. And putting $x=y=0$ in functional equation we get $f(0)=f(0)f(0)$. Since $f(0)>0$ it follows that $f(0)=1$. Next $f(x) f(-x) =f(0)=1$ and hence if $x>0$ then $f(-x) =1/f(x)>0$. Thus we have proved that $f$ takes only positive values. 
Now consider $0<x<1$ and then we have $$\frac{f(x) - 1}{x}\geq 1\tag{1}$$ Next $f(-x) \geq 1-x$ or $1/f(x)\geq 1-x$ or $$f(x) \leq \frac{1}{1-x}$$ or $$\frac{f(x) - 1}{x}\leq \frac{1}{x}\left(\frac{1}{1-x}-1\right)=\frac{1}{1-x}\tag{2}$$ Combining $(1),(2)$ we get $$1\leq \frac{f(x) - 1}{x}\leq\frac{1}{1-x}$$ Letting $x\to 0^{+}$ we get via Squeeze Theorem $$\lim_{x\to 0^{+}}\frac{f(x)-1}{x}=1\tag{3}$$ And because of the above limit we see that $f(x) \to 1$ as $x\to 0^{+}$. Next we have $$\lim_{x\to 0^{-}}\frac{f(x)-1}{x}=\lim_{t\to 0^{+}}\frac{1-f(-t)}{t}=\lim_{t\to 0^{+}}\frac{f(t)-1}{t}\cdot\frac{1}{f(t)}=1$$ Hence we finally have $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\frac{f(x)-1}{x}=1\tag{4}$$ Using this and the functional equation $f(x+y) =f(x) f(y) $ we can easily prove that $f'(x) =f(x) $ for all $x$. With $f(0)=1$ this uniquely characterizes the exponential function. 
A: For $x \lt -1\,$, obviously $1+x \lt 0 \lt e^x\,$. For $x \ge -1\,$, by Bernoulli's inequality: $$\left(1+\frac{x}{n}\right)^n \ge 1 + n \cdot \frac{x}{n} = 1+x \quad\quad\text{for} \;\;n \ge 1$$
Passing to the limit for $\,n \to \infty\,$ gives $\,e^x \ge 1+x\,$.
A: For the case $x > 0$, the derivative $\frac d{dx}e^x$ is greater than one. By mean value theorem, all positive $x$ satisfy
$$\frac{e^x - 1}{x} > 1\\
e^x > 1 + x$$
For the case $x < 0$, the derivative $\frac d{dx}e^x$ is between $0$ and $1$. By mean value theorem, all negative $x$ satisfy
$$\frac{e^x - 1 }{x} < 1\\
e^x > 1 + x$$
A: Let $f(x)=e^x$.
Hence, $f$ is a convex function and $y=x+1$ is a tangent to the graph of $f$ in the point $(0,1)$.
Indeed, the slop is $e^0=1$ and $y-1=1(x-0)$ gives $y=x+1$.
Thus, $e^x\geq x+1$ for all real value of $x$.
About the minimal $a$, for which the inequality $a^x\geq1+x$ is true for all real $x$.
It's obvious that we need $a>1$, otherwise $x=\frac{1}{2}$ will get a counterexample, and let $x>0$. 
Thus, we need $$x\ln{a}\geq\ln(1+x)$$ or
$$\ln{a}\geq\frac{\ln(1+x)}{x}.$$
Thus, $$\ln{a}\geq\lim_{x\rightarrow0^+}\frac{\ln(1+x)}{x}=1,$$
which gives $a\geq e$ and since for $a=e$ our inequality is proven, we have the answer: $e$.
A: By using induction I have showded that $$1+n ≤ e^n$$
for $n \in \mathbb{N}$
Induction start $$P(1):1+1≤ e^1$$ 
$$2 ≤ e $$ 
Induction Step
$$P(n):1+n≤ e^n$$ Adding plus 1 to both sides
$$n+2 ≤ e^n+1$$ 
We know now that:$$e^n+1≤ e^{n+1}$$ Since we know that multiplying by a number will yield a higher result compared to adding the same number under conditions: Both numbers have to be positive and cannot be zero.
$$P(n+1):n+2≤ e^{n+1}$$ 
$$P(n+1):1+n+1≤ e^{n+1}$$ 
Concluding, we know that $$f(x) = e^{x}$$ will also be greater than $$g(x) = 1+x$$ but only for $x \in \mathbb{N}$
Now in the second part, I will prove that there is only 1 intersection at x=0.We define a function $$f(x) = e^{x},g(x)=1+x,f-g=e^{x}-(1-x)$$ f-g is the vertical distance function. And now we show that at one point the distance will be zero (Our only intersection point).
Solve (f-g)'= 0
$$(f-g)'= e^{x}-1$$
$$(e^{x}=1$$
$$x=ln(1)$$
$$x=0$$
Now evaluate the function (f-g) for the value 0. 
$$(f-g)(0)=0$$
In the first part using induction I have showed that function f is greater than g for $x \in \mathbb{N}$.
In the second part I have showed that there is only 1 intersection point.
Concluding, we can now say that f is greater than g for $x \in \mathbb{R}$.
One part of the proof is still open: the negative values of x. Hint: Transformation of functions;f(x-k),g(x-k). All arguments still valid.
A: Well, The derivative of $1 + x$ is $1$ and the derivative of $e^x$ is $e^x$ so $e^x$ is increasing faster whenever $e^x > 1$.  Which happens whenever $x > 0$.  As $x = 0$, $1+x = 1$ and $e^x = 1$ so they are equal there but for all points of positive $x$, $e^x > 1 + x$ as it has increased more.
On the other hand for negative values of $x$, $e^x <1$ so $e^x$ would be increasing slower.  If $1+x$ where ever equal or greater then $e^x$, then $e^x$ would never have been able to "catch up".  So for negative $x$, $e^x > 1+x$ .
But at $x = 0$, $e^x = 1+x$.
A: The minimum value of $a$ for which $a^{x}\geq1+x$ for all $x$ is $a=e.$ To see this, observe the global minimum value of $y=a^{x}-x-1$ for $1<a\leq{e}$ equals $m(a):=\frac{1-\ln(a)+\ln(\ln(a))}{\ln(a)}$ by the second derivative test. (This inequality is clearly not true if $a\leq{1}.$) We seek to find the smallest value $a\in(1,e]$ such that $m(a)\geq{0}.$ Since $m(\sqrt{e})=\ln(\frac{e}{4})<0,$ $m(e)=0$ and $\frac{dm}{da}=-\frac{\ln(\ln(a))}{a\ln^{2}(a)}>0,$ for all $1<a<e,$ it follows by the horse-track principle that $m(a)<0$ for all $1<a<e.$ So $a=e$ is the smallest.
A: Using Taylor's Theorem or the Mean-Value Theorem, you can show that, for every $x\in\mathbb{R}$, there exists $\xi(x)\in\mathbb{R}$ between $0$ and $x$ such that
$$\exp(x)=1+x+\frac{x^2}{2}\,\exp\big(\xi(x)\big)\geq 1+x\,.$$  The equality holds if and only if $$\frac{x^2}{2}\,\exp\big(\xi(x)\big)=0\,,\text{ or } x=0\,.$$
