A Newton-like identity

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.

• Ups, I probably need new glasses. For the right side, plug in $s_j$ and use the Taylor series of $\ln$ ;) – N. S. Nov 6 '11 at 3:02
• @user9176: why not post an answer? :-) – robjohn Nov 6 '11 at 3:21
• @user9176: I think my English is quite okay for a non-native speaker but until a few days ago I wouldn't have had the slightest clue what "the left side is simple preventing something undesirable to happen" could possibly mean. Anyone outside the US/Schaum-system would be similarly stumped. – t.b. Nov 6 '11 at 3:27
• You'll want to see page 427 of Charalambides's book. – J. M. is a poor mathematician Nov 6 '11 at 3:34
• @t.b.: If it's any consolation, I find the acronyms most of my students use that are related to math just as confusing (I'm also not a native speaker either). And I positively dislike "FOIL" and its many derivations. As far as I can tell, it actually makes students unable to expand products unless they happen to be binomial times binomial... – Arturo Magidin Nov 6 '11 at 5:45

$$\log (1-\xi t) = -\sum_{k=1}^{\infty} \xi^k \frac{t^k}{k}.$$
Summing this identity for the different values of $\xi$ and then taking the exponential gives the desired identity.