Imagine you are given a polynomial in the following form: $$ P_n(x) = \prod_{k=0}^n (x-a_k) $$ And I would like to find a general expression for this polynomial, in the form of: $$ P_n(x)=c_nx^n+c_{n-1}x^{n-1}+\dots+c_1x+c_0 $$ And find a formula for $c_n$.
I've tried to compute the first few cases. For $n = 2$: $$ x^{2-0} - (a_0+a_1)x^{(2-1)}+a_0 a_1x^{(2-2)} \\ \;\\ x^2 - (a_0+a_1)x+a_0 a_1x $$ For $n=3$: $$ x^{3-0} - (a_0+a_1+a_2)x^{3-1} + (a_0 a_1 + a_0 a_2 + a_1 a_2)x^{3-2} - (a_0 a_1 a_2)x^{3-3} $$ For $n=4$: $$ x^{4-0} - (a_0+a_1+a_2+a_3)x^{4-1} + (a_0 a_1 + a_0 a_2 + a_0 a_3 + a_1 a_2 + a_1 a_3 + a_2 a_3)x^{4-2} + (a_0 a_1 a_2 + a_0 a_1 a_3 + a_0 a_2 a_3 + a_1 a_2 a_3)x^{4-3} - (a_0 a_1 a_2 a_3)x^{4-4} $$ The pattern the coeficients are making its easy to see. For example: for $n=4$ the coeficient of $x^2 = x^{4 - 2}$, are the different pairs of roots that you can make with 4 different possibilities (in this case, 4 total roots) without repeating and without taking into account the order (that is, that the pair $(1, 2)=(2, 1)$). The following expression: $$ a_0 a_1 + a_0 a_2 + a_0 a_3 + a_1 a_2 + a_1 a_3 + a_2 a_3 $$ Can associated the to the pairs: $$ \{(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)\} $$ To see the pattern more clearly.
For the coefficient of $x^k = x^{n - m}$, we can affirm, by induction, that it will contain the sum of all differents groups of size $m$, that could be made with $n$ different roots; without repetition, and not taking into account order. Also, I could see that the numbers of terms that accompanies each monomial of $x$, follows a Pascal triangle.
But, I'm stuck. I don't know if this formula is even possible to get. Any hint?