Proof by Induction: $n! > 2^{n+1}$ for all integers $n \geq 5.$ I have to answer this question for my math class and am having a little trouble with it. 
Use mathematical induction to prove that $n! > 2^{n+1}$ for all integers $n \geq 5.$ 
For the basis step: $(n = 5)$
$5! = 120$
$2^{5+1} = 2^6 = 64$
So $120 > 64$, which is true.
For the induction step, this is as far as I've gotten: 
Prove that $(n! > 2^{n+1}) \rightarrow \left((n+1)! > 2^{(n+1)+1}\right)$
Assume $n! > 2^{(n+1)}$
Then $(n+1)! = (n+1) \cdot n!$
After this, I'm stuck. Any assistance would be appreciated. 
Thanks! 

Induction step:
\begin{equation}
n! > 2^{n+1} \implies (n+1)! = n!(n+1) > 2^{n+1}(n+1)>2^{n+1}\;·\;2 = 2^{n+1+1} = 2^{n+2}.
\end{equation}
The last inequality is because $n\geq 5$ implies $n+1 \geq 6 > 2$.
