Prove that for every integer $n \ge 4$ the following inequality holds $n! \gt 2^n$
Is my proof by mathematical induction correct?
Base case: when $n=4$, we have $4! \gt 2^4$, this is $24 \gt 16$, therefore true.
Hypothesis: assume $n=k$ for $n \ge 4$, therefore $k! \gt 2^k$
Induction: Prove for $k+1$, we have to prove $(k+1)! \gt 2^{k+1}$
$$\tag{*}(k+1)! \gt 2^{k+1}$$
$$(k+1) \cdot k! \gt 2^k \cdot 2$$
since $k \ge 4$, this means $(k+1) \gt 2$ for all $k$ and from our assumption we know that $k! \gt 2^k$
Therefore,
$$(k+1) \cdot k! \gt 2^k \cdot 2$$
is true completing the proof.
Is this correct?
As David mentioned, your inductive hypothesis should be that $k!>2^k$.
Also, the logic flow in your inductive step leaves a little to be desired. You want to show that $(k+1)!>2^{k+1}$, by using that $k\geq 4$ and that $k!>2^k$. But the first thing to do is to restate what you want to prove (I put a $*$ next to the equation I refer to).
The natural way, with precisely the ideas you used, would be: since $k!>2^k$ and $(k+1)>2$, we have $$ (k+1)!=k!(k+1)>2^k\times 2=2^{k+1}. $$