A way to calculate e? Define three sequences:
The first sequence is $$n^n: 1,\ 4,\ 27,\ 256,\ 3125,\ 46656, \ldots$$
The second sequence is that of the ratios between adjacent members of the first series, or $$\frac{(n+1)^{n+1}}{n^n}: 4,\ \frac{27}4,\ \frac{256}{27}, \ \frac{3125}{256},\ \frac{46656}{3125},\ldots.$$
The third sequence is the difference between adjacent members of the second sequence, or $$\frac{(n+2)^{n+2}}{(n+1)^{n+1}} – \frac{(n+1)^{n+1}}{n^n}: \frac{11}{4},\ \frac{295}{108},\ \frac{18839}{6912},\ \frac{2178311}{800000},\ \ldots.$$
The third sequence converges toward e, from above, and rather quickly so. Is there a proof or explanation of why this must be so?

This is a continuation of @Mark Bennet's answer in which I will show, why the limit is indeed $e$.
We will concentrate on the term $T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)$.
Firstly, we make a few modifications:
$$
T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)=\\
(n+1)\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^n=\\
(n+2)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^n=\\
\underbrace{\left(1+\frac{1}{n+1}\right)^{n}}_{u_n}+\underbrace{(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n}-\left(1+\frac{1}{n}\right)^n\right)}_{v_n}
$$
As we can see quite easily, $u_n$ converges to $e$. For $v_n$ we use the well known factorization $a^n-b^n=(a-b)\cdot\sum_{r=0}^{n-1}a^rb^{n-1-r}$:
$$
v_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n}-\left(1+\frac{1}{n}\right)^n\right)=\\
(n+1)\left(\frac{1}{n+1}-\frac{1}{n}\right)\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}=\\
-\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}
$$
With the obvious inequalities $\left(1+\frac{1}{n+1}\right)^r≤\left(1+\frac{1}{n}\right)^r$ and $\left(1+\frac{1}{n}\right)^{n-1-r}≥\left(1+\frac{1}{n+1}\right)^{n-1-r}$ we obtain:
$$
\left(1+\frac{1}{n+1}\right)^{n-1}=\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n+1}\right)^{n-1-r}≤-v_n\\
≤\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}=\left(1+\frac{1}{n}\right)^{n-1}
$$
Both the upper and the lower bound of $-v_n$ converge to $e$, so by the squeeze theorem, $v_n$ converges to $-e$. Therefore:
$$
\lim_{n\to\infty}T_n=\lim_{n\to\infty}u_n+v_n=\lim_{n\to\infty}u_n+\lim_{n\to\infty}v_n=e-e=0
$$
This completes the answer of Mark Bennet with sufficient precision and we can conclude that your limit indeed is $e$.
