Newton sums (Newton identities for power sums over the roots) proof question

You have by FTA and with the roots $z_1,...,z_n$ of the given polynomial $$ p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=a_n(x-z_1)...(x-z_n). $$ Define $$ q(t)=a_n(1-tz_1)...(1-tz_n)=t^np(1/t). $$ Then for $t$ sufficiently small expand the logarithmic derivatve using geometric series, \begin{align} \frac{q'(t)}{q(t)}&=\frac{d}{dt}\ln|q(t)| \\ &=\frac{d}{dt}(\ln|a_n|+\ln|1-tz_1|+...+\ln|1-tz_n|) \\ &=-\frac{z_1}{1-tz_1}-...-\frac{z_z}{1-tz_z} \\ &=-\sum_{k=0}^\infty z_1^{k+1}t^k-...-\sum_{k=0}^\infty z_n^{k+1}t^k \\ &=-(P_1+tP_2+t^2P_3+t^3P_4+...) \end{align} Now multiply out the denominator and compare coefficients in \begin{multline} 0=\underbrace{(a_{n-1}+2a_{n-2}t...+(n-1)a_1t^{n-2}+na_0t^{n-1})}_{=q'(t)} \\ +\underbrace{(a_n+a_{n-1}t+...+a_1t^{n-1}+a_0t^n)}_{=q(t)}(P_1+tP_2+t^2P_3+t^3P_4+...) \end{multline} to read off the Newton identities \begin{align} 0&=~~a_{n-1}+a_nP_1\\ 0&=2a_{n-2}+a_{n-1}P_1+a_nP_2\\ 0&=3a_{n-2}+a_{n-2}P_1+a_{n-1}P_2+a_nP_3\\ &\vdots\\ 0&=na_{0}+a_{1}P_1+...+a_{n-1}P_{n-1}+a_nP_n\\[1em] 0&=\qquad a_{0}P_1+a_{1}P_2+...+a_{n-1}P_{n}+a_nP_{n+1}\\ 0&=\qquad a_{0}P_2+a_{1}P_3+...+a_{n-1}P_{n+1}+a_nP_{n+2}\\ \end{align} etc.

To be honest, I don't understand Newton's sums well enough to explain them, but what I've figured out is that the equation $$a_nP_k+a_{n-1}P_{k-1}+\ldots +a_0P_{k-n}=0$$ only makes sense when $k\geq n$. By the way, the proof you posted doesn't mention this, but $$P_0=x_1^0+x_2^0+\dots+x_n^0=n$$ In the case that $k < n$ the equation you should actually use is $$a_nP_k+a_{n-1}P_{k-1}+\ldots+a_{n-k+1}P_1+k*a_{n-k}$$ If you want to see the proof for this second equation, then check out this article. It's a pretty rough read, but I couldn't find anything better.