A Newton-like identity

Question

Let $\xi_1, \xi_2, \cdots, \xi_n$ be indeterminates. Define the following indeterminates: $$s_k := \sum\limits_{i=1}^n\xi_i^k, 1\le k <\infty ,$$ $$\sigma_k := \sum\limits_{1\le i_1<i_2<\cdots<i_k\le n}\xi_{i_1}\xi_{i_1}\cdots\xi_{i_k}, 1\le k <\infty.$$

How to show $$ \prod\limits_{i=1}^n(1-\xi_it)=1-\sigma_1t+\sigma_2t^2-\cdots+(-1)^n\sigma_nt^n=\exp\left(-\sum\limits_{j=1}^\infty s_j\frac{t^j}{j}\right)?$$

Thanks.

Answer 1

The power series of the logarithm gives

$$\log (1-\xi t) = -\sum_{k=1}^{\infty} \xi^k \frac{t^k}{k}.$$

Summing this identity for the different values of $\xi$ and then taking the exponential gives the desired identity.