jmc

First, we factor $5!$ into primes: $$\begin{array}{rl}5!& =5\cdot 4\cdot 3\cdot 2\cdot 1\\ & =2^3\cdot 3\cdot 5.\end{array}$$ Then, we factor $\frac{8!}{3!}.$ $$\begin{array}{rl}\frac{8!}{3!}& =8\cdot 7\cdot 6\cdot 5\cdot 4\\ & =2^6\cdot 3\cdot 5\cdot 7.\end{array}$$ We can find the greatest common factor by taking the lowest exponent of each common prime factor. We get $2^3\cdot 3\cdot 5=\boxed{120}.$