How can I prove by induction that $n!>n^{\frac n2}$ for $n>2$? First I've checked it by putting $n=3$.

Then, I've assumed it to be true for $n=k$ to get $k!&gt;k^{\frac k2}$.

Then, putting $n=k+1$, I get $(k+1)!=k!(k+1)&gt;k^{\frac k2}(k+1)$. Now, I'm stuck and couldn't proceed further.
It remains to show that $(k^{k/2})(k+1)\geq (k+1)^{(k+1)/2}$, or $k^k\geq (k+1)^{k-1}$. This is equivalent to $$k\geq\left(1+\frac1k\right)^{k-1}.$$
Does the term on the right look familiar?
It's an easy fact that $${k \choose m} \leq k^m.$$  Then $$(k+1)^{k+1} = k^{k+1} + {k \choose 1}k^k + {k \choose 2}k^{k-1}+ \cdots +1 \leq k^{k+1}+\cdots + k^{k+1} = k^{k+1}(k+1)$$ since there are $k+1$ terms in the sum.  If we take the square root of both sides, this gives your last inequality, so apply the induction hypothesis and you're done.

For the "easy fact", note that $${k \choose m} = \frac{k(k-1)\cdots(k-m+1)}{(k-m)!} &lt; \frac{k^m}{(k-m)!} \leq k^m.$$