Question about the analytic continuation of zeta

Question

Im reading about the analytic continuation of the zeta function. I'll assume that you know the analytic continuation so I'll write in mainfolds,

We prove that $ \varGamma\left(s\right)\zeta\left(s\right)=\intop_{0}^{\infty}\frac{t^{s-1}}{e^{t}-1}dtm$

And we write $ \intop_{0}^{\infty}\frac{t^{s-1}}{e^{t}-1}=\intop_{0}^{1}\frac{t^{s-1}}{e^{t}-1}+\intop_{1}^{\infty}\frac{t^{s-1}}{e^{t}-1} $

Where the latter integral is entire function.

Then, we use lauren expanstion of $ \frac{1}{e^{z}-1} $ in order to continue $ \intop_{0}^{1}\frac{t^{s-1}}{e^{t}-1} $ in an analytic way.

I calculated the first terms in the Laruent expansion and found that $ \frac{1}{e^{z}-1}=\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z-\frac{1}{720}z^{3}+\sum_{k=4}^{\infty}a_{\left(k+1\right)}z^{k} $

So that $$ \intop_{0}^{1}\frac{t^{s-1}}{e^{t}-1}dt=\intop_{0}^{1}\left(\frac{1}{t}-\frac{1}{2}+\frac{1}{12}t-\frac{1}{720}t^{3}+\sum_{k=4}^{\infty}a_{\left(k+1\right)}t^{k}\right)t^{s-1}dt $$

$$ =\intop_{0}^{1}\left(t^{s-2}-\frac{1}{2}t^{s-1}+\frac{1}{12}t^{s}-\frac{1}{720}t^{s+2}+\sum_{k=4}^{\infty}a_{k+1}t^{s+k-1}\right) $$

$$ =\frac{1}{s-1}-\frac{1}{2}\cdot\frac{1}{s}+\frac{1}{12}\frac{1}{s+1}+\sum_{k=4}^{\infty}a_{k+1}\frac{1}{s+k} $$

So that the series we got is an anlytic continuation of the ugly integral, except for the isolated poles

And we get $$ \zeta\left(z\right)\varGamma\left(z\right)=\left[\frac{1}{z-1}-\frac{1}{2}\cdot\frac{1}{z}+\frac{1}{12}\frac{1}{z+1}+\sum_{k=4}^{\infty}a_{k+1}\frac{1}{z+k}+\intop_{1}^{\infty}\frac{t^{s-1}}{e^{t}-1}\right] $$

So

$$ \zeta\left(z\right)=\frac{1}{\varGamma\left(z\right)}\left[\frac{1}{z-1}-\frac{1}{2}\cdot\frac{1}{z}+\frac{1}{12}\frac{1}{z+1}+\sum_{k=4}^{\infty}a_{k+1}\frac{1}{z+k}+\intop_{1}^{\infty}\frac{t^{z-1}}{e^{t}-1}\right] $$

Has only one simple pole at $z=1$.

My questions are:

1. What gives us the right to exchange the summation and itegral order?

2. How can we say that, for example $ \intop_{0}^{1}t^{s-2}dt=\frac{1}{s-1} $ ?

As I see it, $ \intop_{0}^{1}t^{s-2}dt=\frac{1}{s-1}\left(t^{s-1}\right)_{0}^{1}=\frac{1}{s-1}\left(1^{s-1}-0^{s-1}\right) $

What happens with $ 0^{s-1} $? is it defined for $ s \neq 1 $? what is $0^{-6} $ for example? is it well defined?

Thanks in advance. Clarifications will be very helpful to me. (The exchange of the order of summation and integration really bother me and the author did not explain why it is legit in my analysis book)