Prove By Mathematical Induction (factorial-to-the-fourth vs power of two) Prove $(n!)^{4}\le2^{n(n+1)}$ for $n = 0, 1, 2, 3,...$
Base Step: $(0!)^{4} = 1 \le 2^{0(0+1)} = 1$
IH: Assume that $(k!)^{4} \le 2^{k(k+1)}$ for some $k\in\mathbb N$. 
Induction Step: Show $(k+1!)^{4} \le 2^{k+1((k+1)+1)}$
Proof: $(k+1!)^{4} = (k+1)^{4}*(k!)^{4}$ By the definition of factorial. 
$$\begin{align*}
(k+1)^{4}*(k!)^{4} &\le (k+1)^{4}*2^{k(k+1)}\\
&\le (k+1)^{4}*2^{(k+1)((k+1)+1)}
\end{align*}$$ 
by the IH. 
That is as far as I have been able to get at this point...Please Help! Any suggestions or comments are greatly appreciated.  

You might have to do a secondary proof by induction to prove this. What you want to show is the following (I hope my LaTeX renders properly): $((n+1)!)^4 \leq 2^{(n+1)(n+2)}$. We know from the inductive hypothesis that $(n!)^4 \le 2^{n(n+1)}$. Multiplying on both sides by $(n+1)^4$, we have $((n+1)!)^4 \leq (n+1)^4 2^{n(n+1)}$. For the proof to be complete, you only need to show that $(n+1)^4 2^{n(n+1)} \leq 2^{(n+1)(n+2)}$. This can be done easily with a secondary proof by induction after doing a little bit of rearranging of terms in the inequality.
