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It seems to me that you don't need to know that $\zeta(\sigma + it)$ is proportional to $(\sigma+it-1)^k$ for some $k$, you just need to know that, if it vanishes, then it is $O(\sigma+it-1)$. By Taylor's theorem, that will follow if you know that $\zeta(\sigma+i t)$ is a $C^2$ function.

Write $$\zeta(s) = \frac{1}{s-1} + \sum_{n=1}^{\infty} \left( \frac{1}{n^s} - \frac{1}{(s-1) n^sn^{s-1}} + \frac{1}{(s-1)(n+1)^s} frac{1}{(s-1)(n+1)^{s-1}} \right).$$ For $\sigma>1$, this is equal to the standard $\zeta$ by easy manipulations with absolutely convergent series; for $\sigma \in (0,1]$ you can take it as the definition of $\zeta$. Since you don't have analytic continuation, you don't know that this is the "best" extension of $\zeta$ to the critical strip, but that's OK.

Differentiating term by term gives you a series which converges uniformly on compact subsets of $\{ \sigma+it : \sigma>0, \ \sigma+it \neq 1 \}$ Hence $\zeta'$ is represented on this domain by the derivative of this series. The same thing happens when you differentiate again. So $\zeta''$, being the uniformly convergent sum of continuous functions, is continuous and $\zeta$ is $C^2$.
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It seems to me that you don't need to know that $\zeta(\sigma + it)$ is proportional to $(\sigma+it-1)^k$ for some $k$, you just need to know that, if it vanishes, then it is $o(\sigma+it-1)$. O(\sigma+it-1)$. By Taylor's theorem, that will follow if you know that $\zeta(\sigma+i t)$ is a $C^2$ function.

Write $$\zeta(s) = \frac{1}{s-1} + \sum_{n=1}^{\infty} \left( \frac{1}{n^s} - \frac{1}{(s-1) n^s} + \frac{1}{(s-1)(n+1)^s} \right).$$ For $\sigma>1$, this is equal to the standard $\zeta$ by easy manipulations with absolutely convergent series; for $\sigma \in (0,1]$ you can take it as the definition of $\zeta$. Since you don't have analytic continuation, you don't know that this is the "best" extension of $\zeta$ to the critical strip, but that's OK.

Differentiating term by term gives you a series which converges uniformly on compact subsets of $\{ \sigma+it : \sigma>0, \ \sigma+it \neq 1 \}$ Hence $\zeta'$ is represented on this domain by the derivative of this series. The same thing happens when you differentiate again. So $\zeta''$, being the uniformly convergent sum of continuous functions, is continuous and $\zeta$ is $C^2$.
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It seems to me that you don't need to know that $\zeta(\sigma + it)$ is proportional to $(\sigma+it-1)^k$ for some $k$, you just need to know that, if it vanishes, then it is $o(\sigma+it-1)$. By Taylor's theorem, that will follow if you know that $\zeta(\sigma+i t)$ is a $C^2$ function.

Write $$\zeta(s) = \frac{1}{s-1} + \sum_{n=1}^{\infty} \left( \frac{1}{n^s} - \frac{1}{(s-1) n^s} + \frac{1}{(s-1)(n+1)^s} \right).$$ For $\sigma>1$, this is equal to the standard $\zeta$ by easy manipulations with absolutely convergent series; for $\sigma \in (0,1]$ you can take it as the definition of $\zeta$. Since you don't have analytic continuation, you don't know that this is the "best" extension of $\zeta$ to the critical strip, but that's OK.

Differentiating term by term gives you a series which converges uniformly on compact subsets of $\{ \sigma+it : \sigma>0, \ \sigma+it \neq 1 \}$ Hence $\zeta'$ is represented on this domain by the derivative of this series. The same thing happens when you differentiate again. So $\zeta''$, being the uniformly convergent sum of continuous functions, is continuous and $\zeta$ is $C^2$.