Find the coefficient of $x^{20}$ in $(x^{1}+⋯+x^{6} )^{10}$ I'm trying to find the coefficient of $x^{20}$ in 
$$(x^{1}+⋯+x^{6} )^{10}$$
So I did this : 
$$\frac {1-x^{m+1}} {1-x} = 1+x+x^2+⋯+x^{m}$$
$$(x^1+⋯+x^6 )=x(1+x+⋯+x^5 ) = \frac {x(1-x^6 )} {1-x} = \frac {x-x^7} {1-x}$$
$$(x^1+⋯+x^6 )^{10} =\left(\dfrac {x-x^7} {1-x}\right)^{10}$$
But what do I do from here ? any hints ? 
Thanks 

Since $(x+x^2+\cdots+x^6)^{10}=x^{10}(1+x+\cdots+x^5)^{10}$ and $1+x+\cdots+x^5=\frac{1-x^6}{1-x},$ we need to find the coefficient of $x^{10}$ in $(\frac{1-x^6}{1-x})^{10}=(1-x^6)^{10}(1-x)^{-10}.$ 
Since $(1-x^6)^{10}(1-x)^{-10} = (1-10x^6+45x^{12}+\cdots) \sum_{m=0}^{\infty}\binom{m+9}{9}x^{m},$ the coefficient of 
$x^{10}$ will be $\binom{19}{9}-10\binom{13}{9}. $
