Prove that ${\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}}$. Which is greater? $\sqrt{n}^{\sqrt{n+1}}$ or $\sqrt{n+1}^\sqrt{n}$
I know that $\sqrt{n}^{\sqrt{n+1}}$ is greater but I tried using induction and I couldn't figure it out. Thanks for the help.

Hint: take logarithms of both sides and approximate for sufficiently large $n$.
$$\log{\sqrt{n}^{\sqrt{n+1}}} = \frac{1}{2} \sqrt{n+1} \log{n}$$
For sufficiently large $n$, this is approximately
$$-\frac{1}{2}\sqrt{n} \log \left(\frac{1}{n}\right)-\frac{1}{4} \sqrt{\frac{1}{n}} \log
   \left(\frac{1}{n}\right)+\frac{1}{16} \left(\frac{1}{n}\right)^{3/2} \log
   \left(\frac{1}{n}\right)+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$
$$\log{\sqrt{n+1}^{\sqrt{n}}} = \frac{1}{2} \sqrt{n} \log{(n+1)}$$
For sufficiently large $n$, this is approximately
$$-\frac{1}{2}\sqrt{n} \log \left(\frac{1}{n}\right)+\frac{1}{2}\sqrt{\frac{1}{n}}-\frac{1}{4}\left(\frac{1}{n}\right)^{3/2}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$
The latter expression is greater than the former when $n > e^2$.  Because the log is monotonic, then
$$\sqrt{n}^{\sqrt{n+1}} > \sqrt{n+1}^{\sqrt{n}}$$
when $n > e^2$.
