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.

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.
