Find the coefficient of $x^{18}$ in $(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $ 
Find the coefficient of $x^{18}$ in $$(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $$

This is the first time coming across a generating function question and am not quite sure how to solve this and am looking for some help, thanks!

Edit: So it looks like this could have been done much easier by factoring first. I'll leave this answer here in the event that this may be useful to someone with a related problem
Well we can get $x^{18}$ in our equation whenever we multiply terms together who's exponents will add to $18$. The left part of the equation will add in $1,2,3,4,5$ so we need to find the coefficients on the left side who's terms are $x^{17}, x^{16}, x^{15}, x^{14}, x^{13}$
Now to find each coefficient, we need to find all the sets of $5$ numbers in $[2,k]$ that sum to our exponents listed prior. I'll do $13$, for example.
$$
13 = 2+2+2+2+5
$$
$$
13 = 2+2+2+3+4
$$
$$
13 = 2+2+3+3+3
$$
Those are (hopefully I missed none), all the sums that sum to our exponent. now since these 13's can be got by multiplying any order of these terms, we need to see how many combinations these have by using the formula $\frac{n!}{k_1!k_2!...k_n!}$ so,
$$
2+2+2+2+5 \rightarrow \dfrac{5!}{4!1!} = 5
$$
$$
2+2+2+3+4 \rightarrow \dfrac{5!}{3!1!1!} = 20
$$
$$
2+2+3+3+3 \rightarrow \dfrac{5!}{2!3!} = 10
$$
So summing this up we get that $5+20+10=35$, therefore the coefficient in front of our term becomes $35x^{13}$. When we finally multiply that out by the right side to get $x^5\cdot35x^{13} = 35x^{18}$. 
Now in order to finish the problem, you must apply this process to $x^{14}, x^{15}, x^{16}, x^{17}$, and sum their respective coefficients. Good luck! 
