Prove that $ n < 2^{n}$ for all natural numbers $n$. [duplicate]

Counting argument:

Let $S$ be a set with $n$ elements. There are $2^n$ subsets of $S$. There are $n$ singleton subsets of $S$. There is at least one non-singleton subset of $S$, the empty subset.

Proof by induction.

Let $n \in \mathbb{N}$.

Step $1.$: Let $n=1$ $\Rightarrow$ $n\lt2^{n}$ holds, since $ 1\lt 2$.

Step $2.$: Assume $ n \lt 2^{n}$ holds where $n=k$ and $k \geq 1$.

Step $3.$: Prove $n \lt 2^{n}$ holds for $n = k+ 1$ and $ k\geq 1$ to complete the proof.

$k \lt 2^{k}$, using step $2$.

$2\times k \lt 2\times2^{k}$

$ 2k \lt 2^{k+1}\quad(1)$

On the other hand, $k \gt 1 \Rightarrow k + 1 \lt k+k = 2k$. Hence $k+1\lt2k\quad(2)$

By merging results (1) and (2).

$k + 1 \lt 2k \lt 2^{k+1}$

$k + 1 \lt 2^{k+1}$

Hence, $ n \lt 2^{n}$ holds for all $ n \in \mathbb{N}$