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.

The function $f(n) = 2^n - n$ is monotone increasing on $(0,\infty)$ and since $f(1) > 0, f(n)> 0\forall\; n\in\mathbb{N}.$
