Second Round

a. We observe that $(n+1)!=(n+1)\cdot n!$, and therefore that $n!\cdot (n+1)!=(n!{)}^2\cdot (n+1)$. That product is a perfect square if and only if $n+1$ is a perfect square, since $(n!{)}^2$ is a perfect square. For $1\le n\le 100$ this is the case for $n=3,8,15,24,35,48,63,80,99$ (perfect squares minus one that are below $100$). $\square$

b. We rewrite the product $n!\cdot (n+1)!\cdot (n+2)!\cdot (n+3)!$ as follows: $$(n!{)}^2\cdot (n+1)\cdot (n+2)!\cdot (n+3)!=(n!{)}^2\cdot (n+1)\cdot ((n+2)!{)}^2\cdot (n+3).$$ Since $(n!{)}^2$ and $((n+2)!{)}^2$ are both perfect squares, the above product is a perfect square if and only if $(n+1)(n+3)$ is a perfect square. However, $(n+1)(n+3)$ cannot be a perfect square. Indeed, suppose that $(n+1)(n+3)=k^2$ were a perfect square. Since $(n+1{)}^2<(n+1)(n+3)<(n+3{)}^2$ we would have $n+1<k<n+3$, so $k=n+2$. This is impossible because $(n+1)(n+3)=(n+2{)}^2-1$, which is not equal to $(n+2{)}^2$. $\square$