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?
 A: Note you are computing $$(n+2)\left(1+\frac 1{n+1}\right)^{n+1}-(n+1)\left(1+\frac 1{n}\right)^n=$$$$=\left(1+\frac 1{n+1}\right)^{n+1}+(n+1)\left(\left(1+\frac 1{n+1}\right)^{n+1}-\left(1+\frac 1{n}\right)^n\right)$$
You should recognise the terms in brackets to powers $n,n+1$ as converging to $e$. The second term - the difference multiplied by $n+1$ doesn't obviously go to zero - at least not immediately. Your analysis suggests that the error term decreases more rapidly than when you have the first term alone - and that in itself is interesting. 
Others will perhaps analyse the rate of convergence, or you can try that yourself.
A: 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 $e$.
A: As $n$ tends to $\infty$, the second sequence behaves like a line with slope $e$:
$$
\frac{(n+1)^{n+1}}{n^n} = (n+1)\underbrace{\left( 1 + \frac{1}{n}\right)^n}_{\approx\ e} \approx en + e.
$$
When you take the difference between adjacent points on a line, you get the slope -- in this case, $e$.
A: Let's look at the error, assuming we know the basic $e$ limit:
$$\left ( 1+\frac1{n} \right )^n = e^{n \log{\left (1+\frac1{n} \right )}} = e^{1-\frac1{2 n} + \frac1{3 n^2}+ \frac1{4 n^3}+O\left (\frac1{n^4}\right )} = e \left [1-\frac1{2 n} + \frac{11}{24 n^2}-\frac{7}{16 n^3}+O\left (\frac1{n^4}\right ) \right ]$$
Then
$$\begin{align} \left ( 1+\frac1{n+1} \right )^{n+1} &= e \left [1-\frac1{2 (n+1)} + \frac{11}{24 (n+1)^2}-\frac{7}{16 (n+1)^3}+O\left (\frac1{n^4}\right ) \right ] \\ &= e \left [1-\frac1{2 n} + \frac{23}{24 n^2}-\frac{89}{48 n^3}+O\left (\frac1{n^4}\right ) \right ] \end{align}$$
Thus, the OP's sequence looks like, for large $n$:

$$\left(1+\frac 1{n+1}\right)^{n+1}+(n+1)\left[\left(1+\frac 1{n+1}\right)^{n+1}-\left(1+\frac 1{n}\right)^n\right] = e + \frac{e}{24 n^2} + O\left (\frac1{n^3}\right )$$

The error decreases more rapidly as one would expect - the $O(1/n)$ term vanishes.  Furthermore, the sequence approaches $e$ from above rather than below, as observed.
ADDENDUM
The original version of this answer was incorrect.  Amazing that nobody downvoted and the wrong answer got 10 upvotes.  It should have been apparent to me that one needs to expand out to $O(1/n^3)$ to get the correct behavior.  I think @robjohn saw this and got to the correct answer first, but he was too polite to mention this in my answer.
A: This is more like an heuristic but globally works to get hints on such defined series behavior :
Let $(U_{n})_{n\in\mathbb N}$ be your $\frac{(n+1)^{n+1}}{(n)^{n}}$ so that $(U_{n})_{n\in\mathbb N}= F(n) $ if $F : x \mapsto e ^{(x+1)\ln(x+1)-x\ln(x)}$
Then your trying to study $(U_{n+1} -U_{n})$ which behave like $F'$. Such an approach lacks rigor but permits to understand well some results on series (for instance Abel transformation). Compute  $F'$ and see that its limit at infinity is $e$.
A: This result was actually published here:
H. J. Brothers, J. A. Knox, New closed-form approximations to the logarithmic constant e, Math. Intelligencer 20 (1998) 25–29.
Using Maclaurin series, the paper offers a detailed explanation of why this and other related formulas work.
Here is a link to the paper:
https://www.researchgate.net/publication/225907741_New_Closed-Form_Approximations_to_the_Logarithmic_Constant_e
