How to prove $(n!)^4\le2^{n^2+n}$? This may sound like a newbie but
question is to show that
$$(n!)^4\le2^{n^2+n} for \quad n=1,2,3...$$
I know it is true for n=1.  $(1!)^4\le2^2$
and
assume it is true for $1<m\le n$           for all $\quad m\in N$
we have to show for m=n+1. 
$((n+1)!)^2\le^? 2^{(n+1)^2+n+1}$
$((n+1)!)^4=(n!)^4.(n+1)^4\le 2^{n^2+n}.(n+1)^4$
so it is enough to show
$(n+1)^4\le4^{n+1}$
it is not true for n=2 but $(2!)^4\le2^{2^2+2}$ is true 
so we can check for $n\ge3$
$(n+1)^4=n^4+4n^3+6n^2+4n+1\le^?4^{n+1}=4(4^n)$
I need to show $4n+1\le4^n$ ,$4n^3\le4^n$,$6n^2\le4^n,n^4\le4^n$ How Can I continue?

I would rather write $$2^{n^2+n}=4^{\frac{n(n+1)}2}=4^14^24^3\cdots 4^n$$
Then looking at the factors individually, it suffices to how that for each $n\geqslant 4$, that $n^4\leqslant 4^n$. Observe the last inequality is false for $n=3$; but your inequality is. Taking $\log$s, this is equivalent to showing that $$\frac{\log n}n\leqslant \frac{\log 4}4$$ for every $n\geqslant 4$. The inequality is obviously true for $n=4$, and if $x>4$ (in fact, if $x>e$) $$\frac{1-\log x}{x^2}<0$$ so $\dfrac{\log n}n$ decreases for $n>3$.
