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 $s$.

*

*For $Re(s) < 0$ : $$F(s) = -\sum_{m=2}^\infty \sum_{n=1}^\infty {s \choose n} m^{s-n} = -\sum_{m=2}^\infty m^s((1+\frac{1}{m})^s-1) = \sum_{m=2}^\infty (m^s-(m+1)^s) = 2^s $$
Also as $N \to \infty$ : $(1+\frac{1}{m})^s-1 = \sum_{n=1}^N {s \choose n} m^{s-n}+ \mathcal{O}(m^{s-N})$ so inverting the double sum is allowed
$$F(s) = -\sum_{n=1}^\infty \sum_{m=2}^\infty  {s \choose n} m^{s-n} = \sum_{n=1}^\infty {s \choose n}(1-\zeta(n-s))$$

*For any $s \not \in \mathbb{N}$, as $n \to \infty$ :  $1-\zeta(n-s) = \mathcal{O}(2^{s-n})$ so that $\sum_{n=1}^\infty {s \choose n}(1-\zeta(n-s))$ converges and is analytic
$\implies$ by analytic continuation
$$\sum_{n=1}^\infty {s \choose n}(1-\zeta(n-s)) = 2^s$$
is true for any $s$ not an integer 
