Proving $\frac{n^n}{e^{n-1}} 2$. I am trying to prove 

$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}} \text{ for all }n > 2.$$ 

Here is the original source (Problem 1B, on page 12 of PDF)
Can this be proved by induction?
The base step $n=3$ is proved: $\frac {27}{e^2} < 6 < \frac{256}{e^3}$ (since $e^2 > 5$ and $e^3 < 27$, respectively).
I can assume the case for $n=k$ is true: $\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}$.
For $n=k+1$, I am having trouble:
\begin{align}
(k+1)!&=(k+1)k!\\&>(k+1)\frac{k^k}{e^{k-1}}\\&=e(k+1)\frac{k^k}{e^{k}} 
\end{align}
Now, by graphing on a calculator, I found it true that $ek^k >(k+1)^k$ (which would complete the proof for the left inequality), but is there some way to prove this relation?
And for the other side of the inequality, I am also having some trouble:
\begin{align}
(k+1)!&=(k+1)k!\\&<(k+1)\frac{(k+1)^{k+1}}{e^{k}}\\&=\frac{(k+1)^{k+2}}{e^k}\\&<\frac{(k+2)^{k+2}}{e^k}.
\end{align}
I can't seem to obtain the $e^{k+1}$ in the denominator, needed to complete the induction proof.

Since several answers have already proven the result via induction, I see no harm in recording an additional proof that does not use induction.
We claim for all $n\ge 1$, $$\int_{0}^{1} (x\log(x))^n \, dx = \frac{(-1)^n n!}{(n+1)^{n+1}}$$
This can be shown using differentiation under the integral sign; set $k=n$ below: $$f(n) = \int_{0}^{1} x^n \, dx = \frac{1}{n+1} \implies \frac{d^k}{dn^k} f(n) = \int_{0}^{1} x^n (\log(x))^k \, dx = \frac{(-1)^{k} k!}{(n+1)^{k+1}}$$
Since $x \log(x)$ does not change sign, we have $$\int_{0}^{1} |x\log(x)|^n \, dx = \frac{n!}{(n+1)^{n+1}}$$ From this expression, we see that the given inequality is equivalent to $$\frac{1}{(n+1)e^n} < \int_{0}^{1} |x\log(x)|^n \, dx < \frac{1}{e^n}$$
A simple computation shows $|x\log(x)|^n$ attains a maximum of $e^{-n}$ at $x=e^{-1}$, so the upper bound is evident. To see the lower bound, note $|x\log(x)|^n$ is a concave function, so its graph encloses the triangle with vertices $(0,0)$, $(1,0)$, and $(e^{-n}, e^{-1})$. This triangle has area $e^{-n}/2 > e^{-n}/(n+1)$ since $n>2$.
