Using Jensen's inequality, prove: $$~(1) ~~n!<n^n<(n!)^2~~~~~ and ~~~(2)~~~ e^n\geqslant \frac{n^n}{n!}.$$
Attempt. For (1), since $\log$ is strictly concave, using Jensen: $$\log\frac{1+2+\ldots+n}{n}>\frac{\log1+\log2+\ldots+\log n}{n}$$ equivalently $$\log\frac{\frac{n(n+1)}{2}}{n}>\frac{\log n!}{n}$$ equivalently $$\left(\frac{n+1}{2}\right)^n> n!,$$ and since $\frac{n+1}{2}<n$ (for $n>1$), we get the first part. How can one get the second part?
Is $(2)$ a consequence of $(1)$ (regarding somehow that $\left(1+\frac{1}{n}\right)^n<e$)?
Thank in advance.