proof of Waring’s formula

The following is a proof of the Waring’s formula using formalpower series. We will work with formal power series inindeterminate $z$ with coefficients in the ring$\mathbb{Q}[x_1,\mathrm{\dots },x_n]$. We also need the following equality

$$-\log (1-z)=\sum _{j=1}^{\mathrm{\infty }}\frac{z^j}{j}.$$

Taking log on both sides of

$$1-{\sigma }_{1}z+\mathrm{\dots }+{(-1)}^n{\sigma }_n{z}^n=\prod _{m=1}^n(1-x_{m}z),$$

we get

$$\log (1-{\sigma }_{1}z+\mathrm{\dots }+{(-1)}^n{\sigma }_n{z}^n)=\sum _{m=1}^n\log (1-x_{m}z),$$(1)

Waring’s formula will follow by comparing the coefficients on bothsides.

The right hand side of the above equation equals

$$\sum _{m=1}^n\sum _{j=1}^{\mathrm{\infty }}\frac{x_m^j}{j}{z}^j$$

or

$$\sum _{j=1}^{\mathrm{\infty }}\left(\sum _{m=1}^n{x}_m^j\right)\frac{z^j}{j}$$

The coefficient of $z^k$ is equal to $S_k/k$.

On the other hand, the left hand side of (1) can be writtenas

$$\sum _{j=1}^{\mathrm{\infty }}\frac{1}{j}{({\sigma }_{1}z-{\sigma }_2{z}^2+\mathrm{\dots }+{(-1)}^{n-1}{\sigma }_n{z}^n)}^j.$$

For each $j$, the coefficient of $z^k$ in

$${({\sigma }_{1}z-{\sigma }_2{z}^2+\mathrm{\dots }+{(-1)}^{n-1}{\sigma }_n{z}^n)}^j$$

is

$$\sum _{i_1,\mathrm{\dots },i_n}{(-1)}^{i_2+i_4+i_6+\mathrm{\dots }}\frac{j!}{i_1!\mathrm{\cdots }{i}_n!}{\sigma }_1^{i_1}\mathrm{\cdots }{\sigma }_n^{i_n},$$

where the summation is extended over all $n$-tuple$(i_1,\mathrm{\dots },i_n)$ whose entries are non-negative integers, suchthat

$$i_1+i_2+\mathrm{\dots }+i_n=j$$

$$i_1+2i_2+\mathrm{\dots }+n{i}_n=k.$$

So the coefficient of $z^k$ in the left hand side of (1) is

$$\sum _{j=1}^{\mathrm{\infty }}\sum _{i_1,\mathrm{\dots },i_n}{(-1)}^{i_2+i_4+i_6+\mathrm{\dots }}\frac{(j-1)!}{i_1!\mathrm{\cdots }{i}_n!}{\sigma }_1^{i_1}\mathrm{\cdots }{\sigma }_n^{i_n},$$

or

$$\sum {(-1)}^{i_2+i_4+i_6+\mathrm{\dots }}\frac{(i_1+\mathrm{\dots }+i_n-1)!}{i_1!\mathrm{\cdots }{i}_n!}{\sigma }_1^{i_1}\mathrm{\cdots }{\sigma }_n^{i_n}.$$

The last summation is over all $(i_1,\mathrm{\dots },i_n)\in {\mathbb{Z}}^n$with non-negative entries such that $i_1+2i_2+\mathrm{\dots }+n{i}_n=k$.