From this answer, it implicitly becomes evident that
$$\sum_{k=1}^\infty \frac{kn^k}{(k+n)!} =\frac{1}{(n-1)!}$$
holds for any integer $n$.
I guess that the following general formula holds for any $x>0$:
$$\color{blue}{\sum_{k=1}^\infty \frac{kx^k}{\Gamma(k+1+x)}=\frac{1}{\Gamma(x)}}.$$
I could not find the above identity in the literature of the Gamma function. In fact, known expansions of $\frac{1}{\Gamma(x)}$ are complex.
For $x=1$, see this related question.
Could you prove the guess?
We have to prove $$ \sum_{k=1}^\infty \frac{k x^k \Gamma(x)}{\Gamma(k+1+x)} = 1 . $$ Recall the Beta function $$ \int_0^1 t^{a-1} (1-t)^{b-1}\;dt = B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} $$ The term in our sum is $$ \frac{k x^k \Gamma(x)}{\Gamma(k+1+x)} = \frac{x^k B(k+1,x)}{(k-1)!} $$ since $\Gamma(k+1) = k\Gamma(k)$ and $\Gamma(k) = (k-1)!$. Thus $$ \frac{k x^k \Gamma(x)}{\Gamma(k+1+x)} = \frac{x^k}{(k-1)!}\int_0^1 t^{x-1} (1-t)^k\;dt $$ Let's interchange sum and integral [needs justification] to compute $$ \sum_{k=1}^\infty \frac{k x^k \Gamma(x)}{\Gamma(k+1+x)} = \int_0^1 t^{x-1}\sum_{k=1}^\infty (1-t)^k \frac{x^k}{(k-1)!}\;dt \\ = \int_0^1 t^{x-1} (1-t) x e^{x(1-t)}\;dt = t^x e^{x(1-t)}\big|_{t=0}^1= 1 . $$