User:Chutzpan/Sandbox

$$

2^{ab} - 1 = (2^a - 1)(2^{a(b-1)} + 2^{a(b-2)} + ... + 2^a + 1)

$$

let $$ p(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ... + a_0 = (x - r_1)(x - r_2) \cdots (x-r_n) $$

then $$ a_k = (-1)^{n-k} \sum_{1 \leq i_1 < i_2 < ... < i_k \leq n} \frac{r_1r_2 \cdots r_n}{r_{i_1} r_{i_2} \cdots r_{i_k}} $$