Does the analytic continuation of $\zeta(s)$ for $s\in\mathbb{C} \ \backslash \ \{1\}$ require Cauchy's integral formula?

Question

I've seen Cauchy's integral formula used in proofs for the analytic continuation of the Riemann zeta function to the entire complex plane ($s\neq1$). My question is whether a counterexample can be given that does not require the use of Cauchy's integral formula at all to achieve the above analytic continuation for the zeta function.

Answer 1

A proof using zero theorem:

For $s>1$

$$\zeta(s)=\sum_{n\ge 1}n^{-s}=\sum_{n\ge 1}s\int_n^\infty x^{-s-1}dx$$ $$=s\int_1^\infty \lfloor x\rfloor x^{-s-1}dx= \frac{s}{s-1}+s\int_1^\infty f_1(x)x^{-s-1}dx$$ where $f_1(x)=x-\lfloor x\rfloor$ is one-periodic.

Given $f_n$ one-periodic let $c_n=\int_0^1 f_n(x)dx$ and $f_{n+1}(x)=\int_1^x (f_n(y)-c_n)dy$ which is one-periodic again then $$(s+n-1)\int_1^\infty f_n(x)x^{-s-n}dx= c_n+(s+n-1)(s+n)\int_1^\infty f_{n+1}(x)x^{-s-n-1}dx$$ is analytic for $\Re(s) > -n$.

Whence by induction $\zeta(s)-\frac{s}{s-1}$ is entire.