Why is $\sum_{n\ge1} \binom{s}{n}\left(1 - \zeta(n-s)\right)=2^s$ for all $s \in \mathbb{C}$? Probably an easy question, but I found the following identity that seems true for all $s \in \mathbb{C}$:
$$\sum_{n=1}^{\infty} \binom{s}{n}\big(1 - \zeta(n-s)\big)=2^s$$
Why is this the case? Do similar identities exist for $3^s,4^s,...$?
Additional observation:
For $s \in \mathbb{N}$, only a finite sum up to $s+1$ is required to get the exact power, i.e.:
$$\sum_{n=1}^{s+1} \binom{s}{n}\big(1 - \zeta(n-s)\big)=2^s$$ 
and also:
$$\sum_{n=1}^{s+1} \binom{s}{n}\big(1+\frac{1}{2^{n-s}} - \zeta(n-s)\big)=3^s$$ 
etc.
This trick doesn't seem to work for negative integers or non-integer values of $s$.

\begin{eqnarray*}
\sum_{n=1}^{\infty} \binom{s}{n}(1- \zeta(n-s)) =\sum_{n=1}^{\infty}\sum_{m=2} ^{\infty}\binom{s}{n} \frac{1}{m^{n-s}}
\end{eqnarray*}
Invert the sums & use the binomial theorem 
\begin{eqnarray*}
\sum_{m=2} ^{\infty}\sum_{n=1}^{\infty}\binom{s}{n} \frac{1}{m^{n-s}}=\sum_{m=2} ^{\infty} m^s(1-(1+\frac{1}{m^s}))=\sum_{m=2}^{\infty} m^s -(1+m)^s
\end{eqnarray*}
Telescopicy summy thingy & it equals $2^{s}$ ... as you state. 
So in order to get the sum to start at $m=3$ the original sum would need to be 
\begin{eqnarray*}
\sum_{n=1}^{\infty} \binom{s}{n}(1 +\frac{1}{2^{n-s}}- \zeta(n-s)) =3^{s}
\end{eqnarray*}
Not sure where to go to from here ... 
