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.