Prove by Induction -Inductive Step problem The question is prove by induction that $2^n\le n!$ for all $n\ge 4$
So far I have completed the base case so,
Base case
$n=4$
$ 2^4\le4!$
$16\le24$
Therefore the base case holds
Inductive Step
Assume F(n) is true
$2^{n+1}\le (n+1)!$
$2^{n+1}\le n!(n+1)$
$2\cdot 2^n \le n!(n+1)$ <- I don’t know what do after this to get
New Approach
$2\cdot n!\le n!(n+1)$ $ by F(n) $
$2^{n+1}\le (n+1)!$

When $n\geqslant 4$ by assumption we have
$$2^n \leqslant n! $$
Let's multiply both sides on $2$, which gives
$$ 2^{n+1} \leqslant 2 \cdot n!\quad(1)$$
now, knowing, that for $n\geqslant 4$ we have $2 \lt n+1$, we obtain
$$2 \cdot n! \lt (n+1) \cdot n! = (n+1)! \quad(2)$$
$(1)$ and $(2)$ together gives $ 2^{n+1} \lt (n+1)!$.
