Prove Newton's Identities using the properties of a symmetric polynomial. Apologies if I didn't explain this properly in the title. I understand how we progress from (1) down, but I don't understand how to use the coefficients of EQ 5 to prove Newton's Identities.      
Let $a_1,....,a_N$ be $N$ pairwise different elements in a field, and let
    \begin{equation}
  p(x) = \prod_{u=1}^N (1-xa_u) \tag{1}
 \end{equation}
    Denoting the formal derivative of $p$ by $p'$ the (formal) logarithm derivative of $p$ is given by 
    \begin{equation}
  \frac{p'(x)}{p(x)} = -\sum_{u=1}^N \frac{a_u}{1-xa_u} \tag{2}
 \end{equation}
    and so
    \begin{equation}
  \frac{xp'(x)}{p(x)} = -\sum_{u=1}^N \frac{xa_u}{1-xa_u} = -\sum_{n=0}^\infty x^n S_n \tag{3}
 \end{equation}
    where
    \begin{equation}
  S_n = \sum_{u=1}^N a_{u}^n , n = 1,2,....., \tag{4}
 \end{equation}
$i.e.$
\begin{equation}
  xp'(x) = -p(x)(\sum_{n=0}^\infty x^n S_n) \tag{5}
 \end{equation}
    write
    \begin{equation}
  p(x) = \sum_{n=0}^N x^n (-1)^n \sigma_{n} \tag{6}
 \end{equation}
    where $\sigma_0 = 1$ and where the coefficients $\sigma_n$ for $n \geq 1$ are the symmetric functions of the roots. We have
    \begin{equation}
  \sigma_1 = \sum_{u=1}^N a_u  
 \end{equation}
\begin{equation}
  \sigma_2 = \sum_{u_1 < u_2}^N a_{u_1}a_{u_2} \tag{7}
 \end{equation}
\begin{equation}
  \sigma_3 = \sum_{u_1 < u_2 < u_3}^N a_{u_1}a_{u_2} a_{u_3}
 \end{equation}
By Comparing the coefficients of $x^n$ in $(5)$, prove Newton's identities
\begin{equation}
r\sigma_r = \sum_{i+j=r} \sigma_i \lambda_j
\end{equation}
 A: I will use $e_n$ and $p_n$ for elementary and power-sum symmetric polynomials respectively:
$$ \begin{array}{lll}
 e_k(x_1,x_2,\cdots,x_n) & = & \displaystyle \hskip -0.1in \sum_{1\le i_1<\cdots<i_k\le n} \hskip -0.2in x_{i_{\large1}}x_{i_{\large2}}\cdots x_{i_{\large k}} \\[10pt]
 p_k(x_1,x_2,\cdots,x_n) & = & \hskip 0.17in \displaystyle \sum_{1\le i\le n} x_i^k
\end{array} $$
Then we can define the function
$$ f(T)=\prod_{1\le i\le n}(1-x_iT). $$
Vieta's formulas say this can be expanded as
$$ f(T) = \sum_{k=0}^n (-1)^ke_{k}(x_1,\cdots,x_n)T^k. $$
Differentiate (with respect to $T$) and then multiply by $T$:
$$ Tf'(T)=\sum_{k=1}^n (-1)^k e_k kT^k. \tag{$\star$} $$
We're told that this is part of an identity
$$ Tf'(T)=-f(T)\left(\sum_{j=1}^{\infty} p_jT^j\right).   \tag{5} $$
We can multiply out the polynomials on the right side as
$$ -\left(\sum_{i=0}^n (-1)^ie_iT^i \right)
\left(\sum_{j=1}^{\infty} p_jT^j\right) = -\sum_{i=0}^{\infty}\sum_{j=1}^{\infty} (-1)^ie_ip_j T^{i+j} $$
$$ = \sum_{k=0}^{\infty} \left(\sum_{\substack{i+j=k}} (-1)^{i+1} e_ip_j\right)T^k \tag{$\circ$} $$
For convenience, I've extended the summation from $0\le i\le n$ to $i\ge0$ by stipulating $e_i=0$ for $i>n$; there shouldn't be any summands in that case and the so-called "empty sum" is zero.
In equation $(5)$ we can replace the left side with $(\star)$ and the right side with $(\circ)$:
$$ \sum_{k=1}^n (-1)^k e_k kT^k = \sum_{k=0}^{\infty} \left(\sum_{i+j=k} (-1)^{i+1}e_ip_j\right)T^k. $$
Both sides are the same polynomial so they must have the same coefficients.
The coefficient of $T^r$ on the left and right sides are:
$$ (-1)^r e_r r = \sum_{i+j=r} (-1)^{i+1} e_i p_j $$
Dividing both sides by $(-1)^r$ we get
$$ re_r = \sum_{i+j=r} (-1)^{j+1} e_i p_j $$
The signs are supposed to be alternating; a $(-1)$ is missing from your statement of Newton-Girard. Also keep in mind $i\ge0$ and $j\ge1$ in the summation.
