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.

Proof: We will prove the inequality by induction. Since $e^2 > 5$ and $e^3 < 27$, we have
$$\frac {27}{e^2} < 6 < \frac{256}{e^3}.$$ Thus, the statement for $n=3$ is true. The base step is complete.
For the induction step, we assume the statement is true for $n=k$. That is, assume $$\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}.$$
We want to prove that the statement is true for $n=k+1$. It is straightforward to see for all $k > 2$ that $\left(1+\frac 1k \right)^k < e < \left(1+\frac 1k \right)^{k+1}$; this algebraically implies $$\left(\frac k{k+1} \right)^{k+1} < \frac 1e < \left(\frac k{k+1} \right)^k. \tag{$*$}$$ A separate induction proof for the left inequality of $(*)$ establishes $\left(\frac{k+1}{k+2} \right)^{k+2}<\frac 1e$. We now have
\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+1)^{k+2}}{e^k} \left( \frac{k+2}{k+2} \right)^{k+2} \\
&= \frac{(k+2)^{k+2}}{e^k} \left( \frac{k+1}{k+2} \right)^{k+2} \\
&< \frac{(k+2)^{k+2}}{e^k} \frac 1e \\
&= \frac{(k+2)^{k+2}}{e^{k+1}}
\end{align}
and
\begin{align}
(k+1)! &= (k+1)k! \\
&> (k+1) \frac{k^k}{e^{k-1}} \\
&= (k+1) \frac{k^k}{e^{k-1}} \left(\frac{k+1}{k+1} \right)^k \\
&= \frac{(k+1)^{k+1}}{e^{k-1}} \left( \frac k{k+1} \right)^k \\
&> \frac{(k+1)^{k+1}}{e^{k-1}} \frac 1e \\
&= \frac{(k+1)^{k+1}}{e^k}.
\end{align}
We have established that the statement $$\frac{(k+1)^{k+1}}{e^k}<(k+1)!<\frac{(k+2)^{k+2}}{e^{k+1}}$$ for $n=k+1$ is true. This completes the proof.
