Prove $\ 2^{n-1} \leq n! $ for all $\ n \in \mathbb{N}. $ I think I completed the proof correctly, but mine is different from the textbook solution. Would the following prove the statement adequately? Thank you.
Let $\ P(n) = 2^{n-1} \leq n! $ for all $\ n \in \mathbb{N} $
$\ P(1) = 2^{0} \leq 1 $ is true.
Let $\ P(m) = 2^{m-1} \leq m! $ for all $\ m \in \mathbb{N} $
Suppose $\ P(m) $ is true.
$\ P(m+1) = 2^{m-1+1} \leq (m+1)! $
$\ = 2^{m} \leq (m+1)! $
$\ = 2^{m-1} \cdot 2 \leq (m+1) \cdot m! $
Because we are assuming $\ 2^{m-1} \leq m! $ is true, the inequality depends on $\ 2 \leq m+1. $
$\ 2 \leq m+1 = 1 \leq m, $ which is true for all $\ m \in \mathbb{N}. $
Thus, $\ P(m) $ holds for $\ P(m+1). $
By induction, $\ P(n) $ holds for all $\ n \in \mathbb{N}. $
Yes, your idea is correct but write it properly. Firstly, $2^{0} \leqslant 1!$. Now, suppose, $2^{n-1} \leqslant n!$ for some $n \geqslant 1$. Then, $2^{n} = 2.2^{n-1} \leqslant (n+1).n! = (n+1)!$. Hence, $2^{n-1} \leqslant n!$ for all $n \geqslant 1$ follows from induction principle.