I'm watching this proof for $\zeta(2n)$ on YouTube.
This is what I can understand so far:
$${s\over e^{s} -1} = \sum^{\infty}_{n=0} {\beta_n\over n!} s^n$$
Where $\sum^{\infty}_{n=0} {\beta_n\over n!} s^n$ is meant to represent a Taylor Series for ${s\over e^{s} -1}$.
We then re-wrote ${e^{s} -1}$ as its Taylor Series like so:
$${s\over \sum^{\infty}_{n=1} {1\over n!} s^n} = \sum^{\infty}_{n=0} {\beta_n\over n!} s^n$$
Then just rearranged like so:
$${s =\sum^{\infty}_{n=0} {\beta_n\over n!} s^n \sum^{\infty}_{n=1} {1\over n!} s^n}$$
We then diveded both sides by $s$ like so:
$${1 =\sum^{\infty}_{n=0} {\beta_n\over n!} s^n \sum^{\infty}_{n=1} {1\over n!} s^{n-1}}$$
We then made $\sum^{\infty}_{n=1} {1\over n!} s^{n-1}$ start at $n=0$:
$${1 =\sum^{\infty}_{n=0} {\beta_n\over n!} s^n \sum^{\infty}_{n=0} {1\over (n+1)!} s^{n}}$$
Then using Cauchy's Proof for multiplying two power series, we can equate the previous sums to:
$$1= \sum^{\infty}_{n=0} \sum^{n}_{\mu =0} {\beta_{\mu}\over \mu!} {1\over (n- \mu + 1)!} s^n$$
If we multyply by $(n+1)!$ we then get:
$${\sum^{\infty}_{n=0} \sum^{n}_{\mu =0} {\beta_{\mu}\over \mu!} {(n+1)! \over (n- \mu + 1)!} s^n} $$
Which equals:
$${\sum^{\infty}_{n=0} \sum^{n}_{\mu =0} {(n+1)! \over \mu! (n- \mu + 1)!} \beta_{\mu} s^n} $$
This then equal to:
$${\sum^{\infty}_{n=0} \sum^{n}_{\mu =0} {{n+1}\choose \mu} \beta_{\mu} s^n} $$
We can then divide by $(n+1)!$ and get:
$$1 = {\sum^{\infty}_{n=0} {1\over (n+1)!} \sum^{n}_{\mu =0} {{n+1}\choose \mu} \beta_{\mu} s^n} $$
Now here is where I'm confused. The lecturer who made the video said something about the coefficient in ${{n+1}\choose \mu} {\beta_{\mu} s^n}$ is always $0$.
This last statement lost me completely.
Can someone explain please?