An infinite product for $e^{x}$ This is again from MathWorld. Let $ x > 0$ and define a sequence $a_{n}$ recursively as follows $$a_{1} = \frac{1}{x}, a_{n} = n(a_{n - 1} + 1)$$ Show that $$e^{x} = \prod_{n = 1}^{\infty}\left(1 + \frac{1}{a_{n}}\right)$$ I started as follows
\begin{align}
f(m) &= \prod_{n = 1}^{m}\left(1 + \frac{1}{a_{n}}\right)\notag\\
&= \prod_{n = 1}^{m}\left(\frac{a_{n} + 1}{a_{n}}\right)\notag\\
&= \prod_{n = 1}^{m}\left(\frac{a_{n + 1}}{(n + 1)a_{n}}\right)\notag\\
&= \prod_{n = 1}^{m}\frac{1}{(n + 1)}\prod_{n = 1}^{m}\left(\frac{a_{n + 1}}{a_{n}}\right)\notag\\
&= \frac{1}{(m + 1)!}\frac{a_{m + 1}}{a_{1}}\notag\\
&= x\cdot\frac{a_{m + 1}}{(m + 1)!}
\end{align}
So in effect we need to establish that $b_{n} = a_{n + 1}/(n + 1)! \to e^{x}/x$ as $n \to \infty$. I am not sure how to proceed further. Any help would be appreciated.
Update: As many answers below indicate, the above result is false. If anyone is associated with MathWorld please inform them to fix the problem.
 A: Since we have
$$a_{n+1}=(n+1)(a_n+1)\Rightarrow \frac{a_{n+1}}{(n+1)!}=\frac{a_n}{n!}+\frac{1}{n!},$$
we have for $n\ge 2$$$\frac{a_n}{n!}=\frac{a_1}{1!}+\sum_{k=1}^{n-1}\frac{1}{k!}.$$
Hence, from what you got, we have
$$\begin{align}\prod_{n=1}^{\infty}\left(1+\frac{1}{a_n}\right)&=\lim_{m\to\infty}x\cdot\frac{a_{m+1}}{(m+1)!}\\&=\lim_{m\to\infty}x\left(\frac 1x+\sum_{k=1}^{m}\frac{1}{k!}\right)\\&=\lim_{m\to\infty}\left(1+x\sum_{k=1}^{m}\frac{1}{k!}\right)\\&=\color{red}{1+x(e-1)}\end{align}$$
(The answer I got is not $e^x$, but I cannot find any mistake...)
A: So to continue from where you stop: $$\frac{a_{m+1}}{(m+1)!} = \frac{a_m + 1}{m!} = a_1 + \sum{\frac{1}{m!}} \to \frac{1}{x} + e - 1$$ The result is not showing up as expected. 
However, if $a_n = n(a_1a_{n-1} +1)$ then $$e^x = \prod_{n=1}^{\infty}(1+\frac{1}{a_1a_n})$$ Can be shown the same way.
A: The result presented by the OP is manifestly false and it should be signalled as such to the MathWorld administrators. Aside from this, there are several ways to correct it, among them the following: if
$$a_0 = 1, \quad a_{n+1} = \frac {n+1} x (1 + a_n) ,$$
then
$$ \Bbb e ^x = \prod \limits _{n=0} ^\infty \Big( 1 + \frac 1 {a_n} \Big) , \quad x \ne 0 .$$
