A polynomial coefficient

Question

Let $P(x):=1+a_1x+a_2x^2+\cdots+a_nx^n$, then $$(P(x))^m=1+c_1x+c_2x^2+\cdots+c_{mn}x^{mn},$$ how to find the coefficient $c_j$?

Answer 1

Coefficient of $x_0^{k_0}x_1^{k_1}x_2^{k_2}...x_n^{k_n}$ in $(a_0x_0+a_1x_1+a_2x_2...+a_nx_n)^m$ is $$\frac {m!}{k_0!k_1!k_2!...k_n!} a_0^{k_0}a_1^{k_1}a_2^{k_2}...a_n^{k_n}$$ if $ k_0+k_1+k_2+...+k_n=m$, else it is zero. Now,you have $1,x,x^2,x^3,....,x^n$ in place of $x_0,x_1,x_2,...,x_n$. For $x^k$,you will have to look for $1^{k_0}x^{k_1}{x^{2k_2}~}...{x^n}^{k_n}=x^k$