The following is a proof of the Waring's formula using formal power series. We will work with formal power series in indeterminate $z$ with coefficients in the ring $\mathbb{Q}[x_1,\ldots,x_n]$ . We also need the following equality$$ -\log(1-z) = \sum_{j=1}^\infty \frac{z^j}{j}.$$

Taking log on both sides of$$ 1 - \sigma_1z+\ldots + (-1)^n \sigma_n z^n = \prod_{m=1}^n(1-x_mz),$$ we get \begin{equation} \log(1 - \sigma_1z+\ldots + (-1)^n \sigma_n z^n) = \sum_{m=1}^n \log(1-x_mz), \label{eq} \end{equation}Waring's formula will follow by comparing the coefficients on both sides.

The right hand side of the above equation equals$$ \sum_{m=1}^n \sum_{j=1}^\infty \frac{x_m^j}{j}z^j$$ or$$ \sum_{j=1}^\infty \left( \sum_{m=1}^n x_m^j \right) \frac{z^j}{j}$$ The coefficient of $z^k$ is equal to $S_k/k$ .

On the other hand, the left hand side of () can be written as$$ \sum_{j=1}^\infty \frac{1}{j}(\sigma_1z-\sigma_2z^2+\ldots+(-1)^{n-1} \sigma_n z^n)^j.$$ For each $j$ , the coefficient of $z^k$ in$$(\sigma_1z-\sigma_2z^2+\ldots+(-1)^{n-1} \sigma_n z^n)^j$$ is$$\sum_{i_1,\ldots,i_n} (-1)^{i_2+i_4+i_6+\ldots} \frac{j!}{i_1!\cdots i_n!}\sigma_1^{i_1} \cdots \sigma_n^{i_n},$$ where the summation is extended over all $n$ -tuple $(i_1,\ldots,i_n)$ whose entries are non-negative integers, such that