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.

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.
