<s>[INST] Prove $n < 2^n$ for all $n \geq 0$ using induction. Please verify for me?
Base case: $n = 0 $
$0 < 2^0$
$0 < 1$. This is true.
Inductive step: Suppose $n \geq 0$. Assume $P(k)$ is true if $k = n$. We must deduce that $P$ holds for $k+1$.
$n < 2^n$
$n - 2^n < 0$
$(n+1) - 2^{n+1}$
$n+1 - (2^n \times 2)$
$n+1 - (2^n + 2^n)$
$n - (2^n + 2^n) + 1$
$n - 2^n - 2^n + 1$
$(n - 2^n) - (2^n-1)$
We know $(n - 2^n) < 0$, that is to say $(n-2^n)$ is negative.
For all values of $n \geq 0, (2^n-1)$ is always a non-negative number.
A negative number minus a non-negative number is always negative, so
$(n - 2^n) - (2^n-1) < 0$ is true.
$(n - 2^n) < (2^n-1)$.
$n < (2^n + 2^n) -1$
$(n+1) < 2^{n+1}$
So $n < 2^n$ for all $n \geq 0$.
 [/INST] There is an easier way. If $n<2^n$ for some $n$, then $$n+1<2^n+1 \leq 2^n+2^n=2^{n+1}$$.
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