Proof of Newton Girard formula symmetric polynomials Newton Girard formula states that for $k>2$:
\begin{equation}
p_k=p_{k-1}e_1-p_{k-2}e_2+\cdots +(-1)^{k}p_1e_{k-1}+(-1)^{k+1}ke_{k}
\end{equation}
where $e_i$ are elementary symmetric functions and $p_0=n$ with $p_k=x_1^k+\cdots+x_n^k$.
I am using induction to prove this result. I am stuck at the inductive step, that is to show:
\begin{equation}
p_{k}e_1-p_{k-1}e_2+\cdots +(-1)^{k+1}p_1e_k+(-1)^{k+2}(k+1)e_{k+1}= x_1^{k+1}+\cdots+x_n^{k+1}
\end{equation}
I am not able to see how I can use my inductive hypothesis in the left hand side of the above expression.
 A: By  way   of  enrichment  here  is  an   alternate  formulation  using
cycle indices.

Recall that the  OGF of the cycle index $Z(P_n)$  of the unlabeled set
operator $\mathfrak{P}_{=n}$ is given by
$$G(w) = \sum_{n\ge 0} Z(P_n) w^n =
\exp\left(\sum_{q\ge 1} (-1)^{q+1} a_q \frac{w^q}{q}\right).$$
Differentiating we obtain
$$G'(w) = \sum_{n\ge 0} (n+1) Z(P_{n+1}) w^n =
G(w) \left(\sum_{q\ge 1} (-1)^{q+1} a_q w^{q-1}\right).$$
Extracting coefficients we thus have
$$[w^n] G'(w) = (n+1) Z(P_{n+1}) =
\sum_{q=1}^{n+1} (-1)^{q+1} a_q Z(P_{n+1-q})$$
This is apparently due to Lovasz. 

Substitute the cycle indices with the variables $X_1$ to $X_m$ to get
$$(n+1) Z(P_{n+1})(X_1+\cdots+X_m) \\=
\sum_{q=1}^{n+1}  (-1)^{q+1}  (X_1^q+\cdots+X_m^q)
Z(P_{n+1-q})(X_1+\cdots+X_m)$$
This yields
$$(n+1) e_{n+1}(X_1,\ldots,X_m) =
\sum_{q=1}^{n+1}  (-1)^{q+1}  p_q(X_1,\ldots,X_m)
e_{n+1-q}(X_1,\ldots,X_m).$$
Now a choice of variable names yields the result.
 Remark. The  identity for  $G(w)$ follows  from the  EGF  for the
labeled species  for permutations where  all cycles are marked  with a
variable indicating length of the cycle.

This yields
$$\mathfrak{P}
\left(A_1 \mathfrak{C}_{=1}(\mathcal{W})
+ A_2 \mathfrak{C}_{=2}(\mathcal{W})
+ A_3 \mathfrak{C}_{=3}(\mathcal{W})
+ \cdots \right).$$
Translating to generating functions we obtain
$$G(w) = \exp\left(a_1 + a_2 \frac{w^2}{2} 
+ a_3 \frac{w^3}{3} + \cdots\right).$$
The fact that $$Z(P_n) = \left.Z(S_n)\right|_{a_q := (-1)^{q+1} a_q}$$
then confirms the claim.
