trying to prove: $0<\left|e-S_{n}\right|<\frac{3}{(n+1)!} $ I'm trying to  prove that for every $n\in\mathbb{N}
 $  this inequality holds:
$0<\left|e-S_{n}\right|<\frac{3}{(n+1)!}
 $
S.T:  $S_{n}=1+1+\frac{1}{2!}+\dots+\frac{1}{n!}
 $
By induction:
assuming for $n$ and proving for $n+1$:
$\left|e-S_{n+1}\right|=\left|e-S_{n}-\frac{1}{(n+1)!}\right|=\left|(e-S_{n})+(-\frac{1}{(n+1)!})\right|<|e-S_{n}|+\left|-\frac{1}{(n+1)!}\right|<\frac{3}{(n+1)!}+\left|-\frac{1}{(n+1)!}\right|
 $
but how can i continue from this stage?
edit note:
i can't use the fact that:
$lim_{n\rightarrow\infty}S_{n}=e
 $
thank you.

Hint:
$$
\begin{align}
(n+1)!\,(e-S_n)
&=\sum_{k=1}^\infty\frac{(n+1)!}{(n+k)!}\\
&\le\sum_{k=1}^\infty\frac1{(n+2)^{k-1}}\\
&=\frac{n+2}{n+1}
\end{align}
$$

Doing this without the series representation seems difficult. We could use this answer to show that
$$
\left[\lim_{n\to\infty}\left(1+\frac1n\right)^n\right]^x=\sum_{n=0}^\infty\frac{x^n}{n!}
$$
