For $n\in\mathbf{N}^*$, write $x_n:=\sum_{k=0}^n\frac{1}{k!}$. Then $(x_n)$ is a bounded, increasing sequence.
Fix some $m\in\mathbf{N}^*$. For all $m\leq n$, set $$y_{mn}:=1+\sum_{k=1}^m\frac{1}{k!}\left(\prod_{j=1}^k1-\frac{k-j}{n}\right).$$ Then, we have $y_{mn}\leq x_m$ for all $n\geq m$. How can I show that $(y_{mn})_{n\geq m}$ converges to $x_m$ for all $m\in\mathbf{N}^*$?
Edit:
Should I try to show that $x_m$ is the supremum of $(y_{mn})_{n\geq m}$ (this would work since this sequence is increasing)? Or is it better to directly prove it using $\varepsilon$ method? I would appreciate a hint.
Note that $$\prod_{j=1}^k \,\left(1- \dfrac{k-j}{n} \right) = e\,^{\sum_{j=1}^k\, \log \,\left(1- \dfrac{k-j}{n} \right)} $$
As $n \to \infty$, the exponent approaches $0$. Hence $y_{nm} \to 1 + \sum_{k=1}^m \, \dfrac{1}{k!}$.