I have to prove that $n<2^n$ by using the principle of well-order. What I have done so far is, let $P(n) := n<2^n$ and $C= \lnot P(n) = \{ n | n \ge 2^n \}$. Asumme $C$ is not empty, thus $\exists c \in C$.
Now, for $n=0$, $2^0 = 1$ and since $0 < 1$, $0 \notin C$. Which makes $c-1 \in \mathbb{N}$. Here I'm stuck. If I develop the inequality, I have something like:
$$ c-1<2^{c-1} $$ $$ c<2^{c-1}+1 $$ $$ 2c<2^c+2 $$
But then I don't know where to move from here. Any help or pointer will be greatly appreciated.
If $C\neq\emptyset$, then it has a smallest element $c$, which cannot be $0$, since $0<2^0$. And it cannot be $1$, since $1<2^1$.
On the other hand, since $c-1<c$, $c-1<2^{c-1}$. But then$$c=(c-1)+1\leqslant2(c-1)<2\times2^{c-1}=2^c,$$which is impossible, since $c\in C$.