What's the coefficient of the $m^4n^4$ term in the expansion of $(m+n)^8$?
To find the coefficient of the $m^4n^4$ term, I need to find the value of $i$ that makes $k-i = 4$ and $i = 4$. Since $k = 8$ in this case, I can see that $i = 4$ works, and it's the only value of $i$ that works. Therefore, the coefficient of the $m^4n^4$ term is $\binom{8}{4}$. I can use the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ to compute this binomial coefficient. I get $\binom{8}{4} = \frac{8!}{4!4!} = \frac{8\times 7 \times 6 \times 5}{4\times 3 \times 2 \times 1} = 70$.

