Let $\Lambda_k=\sum_{n=0}^{\infty} \frac{n^k}{n!}$. I was practicing evaluating exponential series, and I encountered the following few series $\Lambda_2=2e, \Lambda_3=5e$ and $\Lambda_4=15e$ which can be proven by using the taylor series expression for $\exp(x)=\sum_{n=0}^{\infty}x^n/n!$. But doing this all over may prove challenging in an exam situation. So I tried generalizing the formula for the exponent of $n$ as $k$.
My Attempt:
$$\begin{aligned}\sum_{n=0}^{\infty}\dfrac{n^k}{n!}&=\sum_{n=1}^{\infty}\dfrac{n^{k-1}-1}{(n-1)!}+\sum_{n=1}^{\infty}\dfrac{1}{(n-1)!}\\ &=\left(\sum_{n=2}^{\infty}\left(\dfrac{\sum_{i=0}^{k-2}n^i}{(n-2)!}\right)\right)+e \end{aligned}$$
I'm not sure how to proceed next. A hint in the right direction would be appreciated. Thanks
One way to proceed is to note that for each $j$, if we define $$\phi_j(n)=n(n-1)\cdots(n-j+1)$$ (that is $\phi_j(n)=j!\binom nj$) then $$\sum_{n=0}^\infty\frac{\phi_j(n)}{n!}=\sum_{n=j}^\infty\frac1{(n-j)!}=e.$$ For example, $$n^4=\phi_4(n)+6\phi_3(n)+7\phi_2(n)+\phi_1(n)$$ and so $$\sum_{n=0}^\infty\frac{n^4}{n!} =\sum_{n=0}^\infty\frac{\phi_4(n)+6\phi_3(n)+7\phi_2(n)+\phi_1(n)}{n!} =(1+6+7+1)e=15e.$$ In general then $$\sum_{n=0}^\infty\frac{n^k}{n!} =e\sum_{j=0}^k a_{k,j}$$ where $$n^k=\sum_{j=0}^ka_{k,j}\phi_j(n).$$ Now, there is a combinatorial interpretation for these "connecting coefficients" $a_{k,j}$.