Given $S_n = 1 + \sum_{k=1}^n {1\over k!}$ prove that $e-S_n \le \frac{n+2}{n!(n+1)^2}$ 
Let $S_n$ be a sequence defined by:
  $$
S_n = 1 + \sum_{k=1}^n {1\over k!}
$$
  Prove that:
  $$
e - S_n \le \frac{n+2}{n!(n+1)^2}
$$

This problem comes in the limits section so i may use anything before the definition of a derivative. I've started with the following. Define:
$$
L_n = \left(1+{1\over n}\right)^n
$$
Rewrite $L_n$ as follows:
$$
\begin{align}
L_n &= \left(1+{1\over n}\right)^n \\
&= 1+ \sum_{k=1}^n {n\choose k}{1\over n^k} \\
&= 1 + \sum_{k=1}^n\frac{\overbrace{n\cdot(n-1)\cdot(n-2)\cdots(n-(k-1))}^{n - k + 1 
 \text{ times}}}{k!n^k}
\end{align}
$$
I've then tried to consider the difference between $L_n$ and $S_n$:
$$
L_n - S_n = {1\over 2!}\left(1-{1\over n}\right) - {1\over 2!} + {1\over 3!}\left(1-{1\over n}\right)\left(1-{2\over n}\right) - {1\over 3!} + \cdots \\
= {1\over 2!}\left(1-{1\over n} - 1\right) + {1\over 3!}\left(\left(1-{1\over n}\right)\left(1-{2\over n}\right) - 1\right) + \cdots
$$
Consider each parentheses. I've tried defining a sequence such that:
$$
\forall n \ge 2:a_n = {1\over n!}\left(\prod_{k=1}^{n-1}\left(1-{k\over n}\right) - 1\right)
$$
This sequence seems to always be less than $0$. So we have that:
$$
\forall n\in \Bbb N :a_n \le 0
$$
Since $a_n$ is involved in $L_n - S_n$ we may also conclude that it is also less than $0$, namely:
$$
L_n - S_n = \sum_{k=2}^na_n \le 0
$$
But at the same time:
$$
\frac{n+2}{n!(n+1)^2} \ge 0
$$
So by this we have that:
$$
L_n - S_n \le e - S_n \le 0 \le \frac{n+2}{n!(n+1)^2}
$$
Now I don't see how to proceed. Should I bound the difference from another side? It feels like all i've done so far (above) is not even a usable argument for this problem, so could you please help me prove what's in the problem statement or point to the right direction.
I've made a Visualization of what is written above, perhaps that would be helpful.
 A: We will use the following limit. If $S_n=\sum\limits_{k=0}^{n}\frac{1}{k!}$ then
$$
\lim\limits_{n\rightarrow\infty} S_n = e.
$$
This limit can be derived from $\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{n}\right)^n = e$.
Now we will prove that for any $m\geq 1$ inequality $S_{n+m}\leq S_n+\frac{n+2}{(n+1)(n+1)!}$ holds. Indeed,
$$
S_{n+m}-S_n=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots+\frac{1}{(n+m)!}\leq \frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\ldots+\frac{1}{(n+2)^{m-1}}\right)=\frac{1}{(n+1)!}\cdot \frac{1-\frac{1}{(n+2)^m}}{1-\frac{1}{n+2}}<\frac{n+2}{(n+1)(n+1)!}.
$$
From inequality $S_{n+m}\leq S_n+\frac{n+2}{(n+1)(n+1)!}$ when $m$ tend to infinity we get $e\leq S_n+\frac{n+2}{(n+1)(n+1)!}$ as desired.
A: $\begin{array}\\
e-S_n
&= \sum_{k=n+1}^{\infty} {1\over k!}\\
&= \frac1{(n+1)!}\sum_{k=n+1}^{\infty} {(n+1)!\over k!}\\
&= \frac1{(n+1)!}\sum_{k=0}^{\infty} {(n+1)!\over (n+1+k)!}\\
&= \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {(n+1)!\over (n+1+k)!}\right)\\
&= \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {1\over \prod_{j=1}^{k}(n+1+j)}\right)\\
&< \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {1\over \prod_{j=1}^{k}(n+2)}\right)\\
&= \frac1{(n+1)!}\left(1+\sum_{k=1}^{\infty} {1\over (n+2)^k}\right)\\
&= \frac1{(n+1)!}\left(1+\frac{1/(n+2)}{1-1/(n+2)}\right)\\
&= \frac1{(n+1)!}\left(1+\frac{1}{n+1}\right)\\
&= \frac1{(n+1)!}\left(\frac{n+2}{n+1}\right)\\
&= \frac{n+2}{(n+1)(n+1)!}\\
&= \frac{n+2}{(n+1)^2n!}\\
\end{array}
$
