Check proof that $\prod_{k=1}^n(1+{1\over a_k})$ is bounded if $a_{n+1} = (n+1)(a_n + 1)$ and $a_1 = 1$, $n\in \mathbb N$ 
Let $n \in \mathbb N$ and:
  $$
\begin{cases}
a_1 = 1 \\
a_{n+1} = (n+1)(a_n + 1)
\end{cases}
$$
  Prove that
  $$
x_n = \prod_{k=1}^n\left(1+{1\over a_k}\right)
$$
  is a bounded sequence.

Obviously $x_n > 0$. So suppose $x_n$ has an upper bound $M$:
$$
0 < x_n < M
$$
Inspecting $a_n$ one may see that the sequence is divergent and $a_n \le n$, so:
$$
\begin{align}
a_{n+1} &> a_n \iff  \\
\iff {1\over a_{n+1}} &< {1 \over a_n}
\end{align}
$$
and
$$
\begin{align}
a_n &\ge n \iff \\
\iff {1\over a_n} &\le {1\over n}
\end{align}
$$
Therefore:
$$
1 + {1 \over a_n} \le 1 + {1\over n}
$$
Now lets compare the products:
$$
\prod_{k=1}^n\left(1 + {1 \over a_k}\right) \le \prod_{k=1}^n\left(1 + {1 \over n}\right)
$$
RHS is the definition of $e$ which may be proved to be bounded in various ways so i will not put it here.
So finally we have:
$$
0 < \prod_{k=1}^n\left(1 + {1 \over a_k}\right) \le \prod_{k=1}^n\left(1 + {1 \over n}\right) < 3 \\
0 < x_n < 3
$$
Which proves that $x_n$ is bounded.
Is the above a valid proof?
 A: No! 
It should be
$$
\prod_{k=1}^n\left(1 + {1 \over a_k}\right) \le \prod_{k=1}^n\left(1 + {1 \over k}\right)=n+1.
$$
By the way, $$x_n=
\prod_{k=1}^n\left(1 + {1 \over a_k}\right)=
\prod_{k=1}^n\frac{a_{k+1}}{(k+1)a_k}=\frac{a_{n+1}}{(n+1)!}.$$
A: Observe that $b_{n+1} = \dfrac{a_{n+1}}{(n+1)!}= \dfrac{a_n}{n!} + \dfrac{1}{n!}= b_n+\dfrac{1}{n!}\implies b_n = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2})+\cdots+(b_2 - b_1)+ b_1 = \dfrac{1}{(n-1)!}+\dfrac{1}{(n-2)!}+\cdots + \dfrac{1}{1!} + 1\implies a_n 
 = n!b_n = n + n(n-1) + n(n-1)(n-2)\cdots + n(n-1)(n-2)\cdot\cdot\cdot1+1\implies x_n = \dfrac{a_{n+1}}{(n+1)!} = b_{n+1}= \dfrac{1}{n!}+\dfrac{1}{(n-1)!}+\cdots+\dfrac{1}{1!}+1< \dfrac{1}{n(n-1)}+\dfrac{1}{(n-1)(n-2)}+\cdots + 1 = \dfrac{1}{n-1}- \dfrac{1}{n}+\dfrac{1}{n-2}-\dfrac{1}{n-1}+\dfrac{1}{n-3}-\dfrac{1}{n-2} +\cdots+\dfrac{1}{2} - \dfrac{1}{3}+\dfrac{1}{1} - \dfrac{1}{2}+1 = 2 - \dfrac{1}{n} < 2$. Thus $x_n < 2, \forall n \ge 1$, hence is bounded. 
A: As duly noted by Micheal $\prod_{k=1}^{n}\left(1+\frac{1}{a_k}\right)=\frac{a_{n+1}}{(n+1)!}$, and by defining $b_n$ as $\frac{a_n}{n!}$ we have $b_1=1$ and
$$ b_{n+1}=\frac{a_{n+1}}{(n+1)!} = \frac{a_n+1}{n!} = b_n+\frac{1}{n!}, $$
hence the sequence $\{b_n\}_{n\geq 1}$ is increasing and convergent to
$$ 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots = \color{red}{e}.$$
It follows that
$$ e = 2\left(1+\frac{1}{4}\right)\left(1+\frac{1}{15}\right)\left(1+\frac{1}{64}\right)\left(1+\frac{1}{325}\right)\left(1+\frac{1}{1956}\right)\cdots $$
The involved sequence is A007526.
