Simple proof $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$

Question

The question of proving $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$ has been posed several times. The answers seem to be overly complex - eg (link).

Here $\zeta(s)=\sum 1/n^s$ for complex $s=\sigma+it$.

Question: Is the following simple argument correct?

1. The magnitude of each summand is $|n^{-s}|=|n^{-\sigma-it}|=n^{-\sigma}$.

2. As $\sigma\rightarrow +\infty$, the magnitude of all the terms in the series $\zeta(s)=\sum1/n^s$ therefore $\rightarrow 0$ except the very first which remains 1.

Therefore $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$.

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EDIT: I have corrected the question to say $\sigma \rightarrow +\infty$.

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