Prove by induction: $n! \ge 2^{(n-1)}$ for any $n \ge 1$ [duplicate]
Solution 1:
Assume $n!\ge 2^{n-1} $ for some $n\ge 1$.
$$(n+1)!=(n+1)n!\ge (n+1)2^{n-1} $$
but $n+1\ge 2$ thus
$$(n+1)2^{n-1}\ge 2^{n-1+1} $$ Done!
Solution 2:
$P(n+1)$ asserts that $(n+1)!\geqslant2^n$. And your are assuming that $n!\geqslant2^{n-1}$. But then$$(n+1)!=(n+1)\times n!\geqslant2\times2^{n-1}=2^n.$$