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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?

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Consider the ratio $\left(\frac{n+1}{n}\right)^4$. This is $\left(1+\frac{1}{n}\right)^4$. For $n\ge 3$, this is $\le \left(1+\frac{1}{3}\right)^4\lt 4$. –  André Nicolas Dec 27 '13 at 0:32

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up vote 10 down vote accepted

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 $n>3$.

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I don't have a better approach, so I'm not contributing a separate answer. I'll note that, using a somewhat different method, the proof comes down to showing that $ \ \log_2 n \ \le \ \frac{n}{2} \ $ , which also fails at $ \ n = 3 \ $ , with equality holding at $ \ n = 2 \ $ and $ \ n = 4 \ $ . –  RecklessReckoner Dec 27 '13 at 4:42

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