Coefficient in a polynomial The coefficient of $x_1^{r_1}...x_k^{r_k}$ in $(x_1 +x_2+...x_k)^n$ is ${n \choose r_1r_2\dots r_k}$. 

How would I go about finding the coefficient of $x_1^{r_1}...x_k^{r_k}$ in $(x_1^2+\dots+x^2_k)^n$.

Thanks!
That's easy : substitute your $x^2_i,\,i\in\{1,2,\dots,k\}$ like this :
$$ \forall 1\leqslant i\leqslant k : x^2_i=y_i $$
Then :
$$ [y_1^{\nu_1}y_2^{\nu_2}\dots y_k^{\nu_k}](y_1+y_2+\dots+y_k)^n = {n \choose {\nu_1,\nu_2,\dots,\nu_k}}, $$
where $[m^2](1-m)^3$ is the coefficient of $m$ in the development of $(1-m)^3$ (that is : $-3$), and $\sum_1^k \nu_i = n$.

Substitute back, and the coefficient of $x_1^{r_1}\dots x_k^{r_k}$ in the expansion of your polynomial is :

$$[x_1^{r_1}x_2^{r_2}\dots x_k^{r_k}](x_1^2+x_2^2+\dots+x_k^2)^n = {n \choose {\nu_1,\nu_2,\dots,\nu_k}},$$
where, for all $i\in\{1,2,\dots,k\}$, we have : $r_i=2\nu_i$
That's easy : substitute your $x^2_i,\,i\in\{1,2,\dots,k\}$ like this :
$$ \forall 1\leqslant i\leqslant k : x^2_i=y_i $$
Then :
$$ [y_1^{\nu_1}y_2^{\nu_2}\dots y_k^{\nu_k}](y_1+y_2+\dots+y_k)^n = {n \choose {\nu_1,\nu_2,\dots,\nu_k}}, $$
where $[m^2](1-m)^3$ is the coefficient of $m$ in the development of $(1-m)^3$ (that is : $-3$), and $\sum_1^k \nu_i = n$.

Substitute back, and the coefficient of $x_1^{r_1}\dots x_k^{r_k}$ in the expansion of your polynomial is :

$$[x_1^{r_1}x_2^{r_2}\dots x_k^{r_k}](x_1^2+x_2^2+\dots+x_k^2)^n = {n \choose {\nu_1,\nu_2,\dots,\nu_k}},$$
where, for all $i\in\{1,2,\dots,k\}$, we have : $r_i=2\nu_i$