For the left-hand equality: You can prove this by induction: The ratio of ${(n+1)^n \over n!}$ to ${n^{n-1} \over (n-1)!}$ is exactly ${(n-1)! \over n!} {(n+1)^n \over n^{n-1}}$ $= {1 \over n}{(n+1)^n \over n^{n-1}}$ $= {(n+1)^n \over n^n}$ $= (1 + {1 \over n})^n$ This is exactly the $n$-th factor of the right-hand side.
For the right-hand inequality: You can do this using the inequality $\ln(1 + x) < x$ for $x > 0$, which can be proven for example by observing that the two functions $\ln(1 + x)$ and $x$ are equal when $x = 0$ while the derivative of $\ln(1 + x)$ is less than the derivative of $x$ (namely $1$) when $x \geq 0$.
Plugging in $x = 1/k$ for any $k$ you get $\ln(1 + {1 \over k}) < {1 \over k}$. Taking $e$ to both sides you get $1 + {1 \over k} < e^{1 \over k}$ which implies $(1 + {1 \over k})^k < e$. Multiplying all of these together from $k = 1$ to $k = n$ gives the inequality you seek.