Show that $0
Show that $0<e-\sum\limits_{k=0}^n\frac{1}{k!}<\frac{1}{n!n}$, where $n>0$.
Hint:
Show that $y_m:=\sum\limits_{k=n+1}^{m+n}\frac{1}{k!}$ has the limit $\lim\limits_{m\to\infty}y_m=e-\sum\limits_{k=0}^n\frac{1}{k!}$ and use that $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$.
If I use the hint then the problem is pretty straightforward.
However, I don't understand why the inequality of the hint is true? A simple example shows that the inequality doesn't hold! There are no further information given.
What am I missing? Is $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$ simply nonsense? Or did the professor forgot to mention some more assumptions?

We have
$$\begin{align}y_m &= \frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots + \frac{1}{(n+1)(n+2)\cdots(n+m)}\right] \\&<\frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots + \frac{1}{(n+1)^m}\right]\end{align}$$
Evaluating the finite geometric sum on the RHS it follows that
$$y_m  < \frac{1}{n!}\frac{\frac{1}{n+1}- \frac{1}{(n+1)^{m+1}}}{1 - \frac{1}{n+1}}= \frac{1}{nn!}\left[1 - \frac{1}{(n+1)^{m}} \right] \underset{ m \to \infty}\longrightarrow 
 \frac{1}{nn!}  $$
