$k! > \frac{k}{2}^\frac{k}{2} \text{ for } k\in N$
I'm not sure how to prove this, is it valid to rewrite it as?:
$(2k)! > k^k \text{ for } k\in N$
Edit: I think I proved it...
$(2k)!=1\cdot2...k\cdot(k+1)...(k+k)>k!k^k$
$(2k)!>k!k^k>k^k$
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1It would be better to define a new variable $m=k/2$ and write it as $(2m)!\gt m^m$ to avoid confusion, but the expressions are equivalent. Now think about the largest $m$ factors on the left. – Ross Millikan Nov 23 '16 at 16:55
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No, it's valid to rewrite it as $(2(k!))^2 > k^k$. Note that $(2(k!))^2$ is very different from and much smaller than $(2k)!$. – Arthur Nov 23 '16 at 16:55