TITLE: Generalized Sophomore's dream. Question about originality
QUESTION [7 upvotes]: A few months ago I derived a beautiful fact:
$$
\sum_{n=k+1}^\infty n^{k-n}=\int_{0}^{1} t^{k-t}dt~~~(*)
$$
for every natural $k$. Generally:
$$
\sum_{n=1}^\infty \frac{a^n}{(n+s)^n}=\int_{-s}^{a-s} \frac{a^t}{(t+s)^t}dt
$$
It is easy to prove. You just need to use the fact that
$$
\frac{1}{(n+c)^n}=\int_{0}^\infty \frac{t^{n-1}}{(n-1)!} e^{-t(n+c)}dt
$$
I know about Sophomore's dream, but even after a long search I didn't find fact $(*)$ in the literature. Please help me and answer, is it original or not?
(+ 1 edition) After some time I derived another generalization (the fact above is just the case with $b=0$):
$$
\sum_{n=1}^\infty \frac{a^{n}}{(n+c)^{n}} \frac{\Gamma(n+b)}{\Gamma(n)(n+c)^b}=\int\limits_{-c}^{a-c} \left ( \mathrm{ln}\frac{a}{t+c}\right )^b\frac{a^t}{(t+c)^t} dt
$$
(where $\Gamma(x)$ denotes the Gamma-function). Other sum is interesting too:
$$
\sum_{n=0}^\infty \frac{a^{n}\Gamma(n+b)\Gamma(n+d)}{n!(n+c)^{n+b}\Gamma(d)}=\int_{0}^\infty \frac{x^{b-1}e^{-x(c-d)}}{(e^x-ax)^{d}} dx
$$

REPLY [6 votes]: Although the OP has stated that he/she has already derived the more general form of the so-called "Sophomore's Dream," I thought that it might benefit others to see the development herein.  So, here we go ...
$$\begin{align}
\int_{-s}^{a-s} \frac{a^t}{(t+s)^t}dt&=a\int_0^1 t^st^{-at}dt \tag1\\\\
&=\sum_{n=0}^{\infty}\frac{(-1)^na^{n+1}}{n!}\int_0^1t^{n+s}\log^nt\,dt \tag2\\\\
&=\sum_{n=0}^{\infty}\frac{(-1)^na^{n+1}}{n!}\frac{(-1)^n}{(n+1)^{n+1}}\int_0^{\infty} t^ne^{\frac{n+1+s}{n+1}x}dx \tag3\\\\
&=\sum_{n=0}^{\infty}\frac{a^{n+1}}{n!(n+1+s)^{n+1}}\int_0^{\infty}t^ne^{-t}dt \tag4\\\\
&=\sum_{n=0}^{\infty}\frac{a^{n+1}}{(n+1+s)^{n+1}}\tag5\\\\
&=\sum_{n=1}^{\infty}\frac{a^{n}}{(n+s)^{n}}\tag6
\end{align}$$
as was to be shown!

NOTES:
$(1)$ 
We enforced the substitution $t \to at-s$
$(2)$ 
We wrote $t^{-at}=e^{-at\log t}$ and used the power series representation for $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$.  We also used the uniform convergence of the power series to justify interchanging the integral and summation.
$(3)$
We enforced the substitution $t\to e^{-t/(n+1)}$.
$(4)$
We enforced the substitution $t\to \frac{n+1}{n+1+s}t$.
$(5)$
We used the integral representation of the Gamma Function $\Gamma (z)=\int_0^{\infty}t^{z-1}e^{-t}dt$, which for $z=n+1$ is $\Gamma (n+1)=n!$.
$(6)$
We shifted the index of summation using $n\to n-1$