Compute $\frac{1}{e}\sum\limits_{n=0}^{\infty}\frac{n^{k}}{n!}$ for $k=0, 1, 2 ... $

Question

Using some matlab (I know it's cheating) I found that: $$k=0 => Result=1$$ $$k=1 => Result=1$$ $$k=2 => Result=2$$ $$k=3 => Result=5$$ $$k=4 => Result=15$$ $$k=5 => Result=52$$ $$k=6 => Result=203$$

This sequence is puzzling...

Answer 1

$$\Large\sum_{n=1}^{\infty}\frac{\mathbb e^{nx}}{n!}=\mathbb e^{\mathbb e^x}-1$$

$$~$$

$$\Large\sum_{n=1}^\infty \frac{n^k}{n!}=\frac{d^k}{dx^k}(\mathbb e^{\mathbb e^x}-1)\huge]_{x=0}$$