How can we show that $n < 2^n$ for all natural numbers $n$? I proved it using calculus or by drawing their graph but I was thinking if there is any simpler way to prove it. Please help me.
Proof by induction >
$P(n) : n < 2^n$ for all $n \in\mathbb{N}$
$P(1) : 1 < 2^1$, i.e.
$1 < 2,$ this is a true statement. 
Now lets assume $P(m)$ is true i.e. $m < 2^m$ . 
So $P(m + 1) : m + 1 < 2^{m + 1}.$ 
Now 
$m < 2 ^ m \Rightarrow 2m < 2^{m + 1} \\
\Rightarrow m+m < 2^{m+1}\\
\Rightarrow m+1 <= m+m < 2^{m+1}\\
\Rightarrow m+1 < 2^(m+1)$
Hence $P(m+1)$ is true. Thus $P(m)$ is true $\Rightarrow P(m+1)$ so by principle of mathematical induction $P(n)$ is true for all $n \in \mathbb{N}.$
So as I said I know the proof using induction hence I wanted to know any other way to prove it.
 A: Since you ask for proof other than that by induction, consider $$2^n=(1+1)^n=\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n} > \binom{n}{1}=n$$ as an application of Binomial theorem.
A: $2^n$ is the cardinality of the power set of a set of $n$ elements.
The power set includes $n$ singletons and the empty set.
A: Summary of comments that should have been answers:


*

*If $n<2^n$ with $n\ge1$ then $n+1\le 2n<2^{n+1}$, so we can induct from $n=1$ onward, and only need to check the base steps $n=0$ and $n=1$;

*A hypothetical function $f$ mapping elements of $\{1,\,\cdots,\,n\}$ to subsets thereof can't satisfy $f(k)=\{m|m\notin f(m)\}$ (as this would lead to the contradiction $k\in f(k)\iff k\notin f(k)$), so there are too many subsets to pair with elements.


A recent comment reveals the first approach is known & not desired.
A: A combinatorial proof:
Note that $$2^n=\sum_{i=0}^n\,\binom{n}{i}=\binom{n}{0}+\binom{n}{1}+\cdots,$$ that $$n=\binom{n}{1}$$ and that all terms are of the sum are positive.
Then the result follows immediately.
A: Let f(X)=X-$2^X$ for all X $\ge$1
$$f'(X)=1-2^X.ln(2)\lt 0$$
f(X) is decreasing function for all X$\ge$1.
Hence $f(X)\lt f(1)=-1\lt 0$
Hence $X\lt 2^X\,\,\,\,\forall X\ge 1$
A: The Bernoulli inequality says that, if $t>-1$ and $n$ is any nonnegative integer, then
$$(1+t)^n\ge 1+nt$$
this inequality can be proved by induction (and is more useful than just proving that $n<2^n$). It is true for $n=0$, because both sides have the value $1$.
Suppose $(1+t)^n\ge 1+nt$, then
$$\begin{align}(1+t)^{n+1}&=(1+t)^n(1+t)\\&\ge(1+nt)(1+t)\\&=1+(n+1)t+nt^2\\&\ge1+(n+1)t\end{align}$$
We used $t>-1$ to write the first inequality; the last one holds because $nt^2\ge0$.
Now apply the inequality to the case $t=1$:
$$2^n=(1+1)^n\ge1+n1=1+n>n$$
A: We can prove this by induction on $n$:
Base case: clearly $n < 2^n$ is true for $n=1$ because $1 < 2^1 = 2$. This proves the base case.
Inductive step: Assume $n < 2^n$ for some $n \in \mathbb{N}$. This implies $2n < 2 \cdot 2^n = 2^{n+1}$. Moreover, $n+1 \leq 2n$ for all $n \geq 1$. So $n+1 \leq 2n < 2^{n+1} \implies n+1 < 2^{n+1}$. This completes the proof. 
A: Induction:
If $k < 2^k$ (which is true for the first few natural $k$) and $k \ge 1$ (which is true for all $k$) then:
$k +1 \le k + k < 2^k + 2^k = 2^{k+1}$ and thus for any natural number this is true for it will be true for the next and there will be none where it isn't true.
.....
In essence, multiplying a number, $n$ by $b \ge 2$ (which is adding $n \ge 1$ to $n$ a positive number of times) results in a larger value then adding $1$ to the number (because adding $1$ is less [or equal] to adding $n$ and the number times done in multiplying are more than the just once of adding $1$).  So $n$ is the result of adding $1$, $n$ times, while $2^n$ is the result of multiplying $2$ $n$ times.  Each multiplication by $2$ must result in a larger value than just adding $1$ would.
A: The function $f(n) = 2^n - n$ is monotone increasing on $(0,\infty)$ and since $f(1) > 0, f(n)> 0\forall\; n\in\mathbb{N}.$
A: By Cantor, for every set $X$, $|\mathcal P(X)| > |X|$. When $X$ is a set with $n$ elements, it follows that $2^n > n$.
