Proving the generating functions version of Newton's identities We are asked to prove the identity in the generating functions version in the Expressing elementary symmetric polynomials in terms of power sums section of the Wikipedia page on Newton's identities
The following is what I am thinking about now:
I have not even seen generating functions before, and since $\infty$ is not a number, I do not know that if it suffices to prove that for any $n$ in N every coefficient of $T^n$ consists. If so, I think maybe it can be done using induction. 
I have not familiar with the exponential generating functions so I am afraid that there are some useful properties of this function which I do not know. If so, could someone tell me what property of this function should I use?
Seeing this question:Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ 
I guess maybe there is an algorithm to translate a given recurrence into generating functions.
But until now I have not found a track to prove it, Could someone please prove it or give some hints? Thanks in advance!
EDIT: The proofs in the link
https://holdenlee.wordpress.com/2010/11/22/newton-sums/ and 
https://yakovenko.files.wordpress.com/2011/04/lecture5symmetry.pdf
are understandable and in generating functions version.
But they are not in exponential generating functions version. I do know that there is a method to translate ordinary generating functions into exponential version. But I cannot understand the principle behind the method and I am afraid that I cannot use that. So is it possible to modify the proofs above to give the desired conclusion? Any hint will be appreciate? Thanks!
 A: What you  need here is the  following statement: the OGF  of the cycle
index $Z(P_k)$  of the  unlabeled set operator  $\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}_{=k}$ is
given by
$$Z(P_k) = [w^k]
\exp\left(\sum_{l\ge 1} (-1)^{l-1} a_l \frac{w^l}{l}\right).$$
This  is known as  the exponential  formula.  To  see this  consider a
repertoire of variables $\sum_q  X_q$ and substitute the corresponding
power sum
$$a_l = \sum_q X_q^l$$
into the formula to get
$$Z(P_k)\left(\sum X_q\right)
= [w^k]
\exp\left(\sum_{l\ge 1} (-1)^{l-1} \sum_q X_q^l \frac{w^l}{l}\right)
\\ = [w^k] \prod_q
\exp\left(\sum_{l\ge 1} (-1)^{l-1} X_q^l \frac{w^l}{l}\right)\\
\\ = [w^k] \prod_q \exp\log(1 + w X_q)
= [w^k] \prod_q (1 + w X_q)$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{Q\subseteq X, |Q|=k} \prod_{X_q\in Q} X_q}$$
by  inspection and  we have  expressed the  elementary  polynomials in
terms of power sums.
Readings. A closely related identity was discussed at this MSE
link  and the proof
was          given         at          this          MSE         link
II.   An  alternate
proof due to Lovasz is included among the links at Wikipedia on cycle
indices.
Example.   The   cycle   index   $Z(P_6)$  for   the   operator
$\textsc{SET}_{=6}$ is given by
$$Z(P_6) = {\frac {{a_{{1}}}^{6}}{720}}-1/48\,a_{{2}}{a_{{1}}}^{4}
\\+1/18\,a_{{3}}{a_{{1}}}^{3}+1/16\,{a_{{1}}}^{2}{a_{{2}}}^{2}
-1/8\,a_{{4}}{a_{{1}}}^{2}-1/6\,a_{{1}}a_{{2}}a_{{3}}
\\+1/5\,a_{{5}}a_{{1}}-1/48\,{a_{{2}}}^{3}
+1/8\,a_{{2}}a_{{4}}+1/18\,{a_{{3}}}^{2}-1/6\,a_{{6}}.$$
