How to prove that $\lim_{n\rightarrow\infty}\left[\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}\right] = e$ I learnt on Wolfram MathWorld that
$$\lim_{n\rightarrow\infty}\left[\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}\right] = e$$
How should I prove this?
Attempt:
$$\begin{align}
\lim_{n\rightarrow\infty}\left[\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}\right] &= \lim_{n\rightarrow\infty}\left[\left(1+\frac{1}{n}\right)^n(n+1)-\left(1+\frac{1}{n-1}\right)^n(n-1)\right]\\
&=\lim_{n\rightarrow\infty}e(n+1)-e(n-1)\\
&= 2e
\end{align}$$
Why did I get a $2e$? Where did I do wrong? Isn't
$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^n=e?$$
 A: The step where you go from $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n(n+1)- \left(1+\frac{1}{n-1}\right)^n(n-1)$ to $\lim_{n\to\infty} e(n+1)-e(n-1)$ is invalid.  You can't evaluate just some pieces of the limit in terms of $n$.  Formally, you're starting with an indeterminate form of type $\infty - \infty$.  
As a simpler example as to why your reasoning doesn't work, consider the indeterminate form $(n+1)-n$, which is identically equal to $1$.  But it can be rewritten to be of the form 
$$
n\left(\frac{n+1}{n}\right) - (n+1)\left(\frac{n}{n+1}\right)
$$
Now both $\left(\frac{n+1}{n}\right)$ and $\left(\frac{n}{n+1}\right)$ converge to $1$, but the limit of $n(1) - (n+1)(1)$ is now $-1$ instead of $1$.
As to how to derive the identity, the Mathworld link suggests that a reasonable approach would be to start with the expression
$$
\left(1 + \frac{1}{x}\right)^x
= e\left[1 - \frac{1}{2x} + \frac{11}{24x^2} - \frac{7}{16x^3} + \frac{2447}{5760x^4} - \dots\right]\, ,
$$
which is derived by using a composition of the identity
$$
x\ln\left(1 + \frac{1}{x}\right)
= 1 - \frac{1}{2x} + \frac{1}{3x^2} - \frac{1}{4x^3} + \dots
$$
with the power series expansion of $e^x$.
A: By Taylor's espansion as $x \to 0$


*

*$\log(1+x)=x-\frac12x^2+o(x^2)$

*$e^x=1+x+o(x)$
we have that
$$\frac{(n+1)^{n+1}}{n^n}=(n+1)\left(1+\frac1n\right)^n=(n+1)e^{n\log\left(1+\frac1n\right)}=$$$$=(n+1)e^{n\left(\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right)\right)}=(n+1)e^{1-\frac1{2n}+o\left(\frac1{n}\right)}=e(n+1)\left(1-\frac1{2n}+o\left(\frac1{n}\right)\right)$$
$$\frac{n^n}{(n-1)^{n-1}}=(n-1)\left(1-\frac1n\right)^{-n}=(n-1)e^{-n\log\left(1-\frac1n\right)}=$$$$=(n-1)e^{-n\left(-\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right)\right)}=(n-1)e^{1+\frac1{2n}+o\left(\frac1{n}\right)}=e(n-1)\left(1+\frac1{2n}+o\left(\frac1{n}\right)\right)$$
and then
$$\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}=en+e-\frac e 2-(en-e+\frac e 2)+o(1)=e+o(1)\to e$$
A: Assuming that the limit exists (which should be proved) we can easily see that it needs to be equal to $e$, indeed by Stolz-Cesaro with $a_n=\frac{n^n}{(n-1)^{n-1}}$ we have
$$a_{n+1}-a_n=\frac{a_{n+1}-a_n}{(n+1)-n} \to L \implies \frac{a_n}{n}=\frac{n^{n-1}}{(n-1)^{n-1}}=\frac{1}{\left(1-\frac1n\right)^{n-1}}\to L=e$$
