jmc

Setting $m=n-1$ in the given functional equation, we get $$\frac{f(n)\cdot f(n)}{f(n-1)}+f(n)=f(n+1),$$for all $n\ge 1.$ Then $$\frac{f(n)}{f(n-1)}+1=\frac{f(n+1)}{f(n)}.$$Let $g(n)=\frac{f(n)}{f(n-1)}$ for $n\ge 1.$ Then $g(1)=\frac{f(1)}{f(0)}=1,$ and $$g(n)+1=g(n+1).$$Then $g(n)=n$ for all $n\ge 1.$ Hence, $$g(n)g(n-1)\cdots g(2)g(1)=\frac{f(n)}{f(n-1)}\cdot \frac{f(n-1)}{f(n-2)}\cdots \frac{f(2)}{f(1)}\cdot \frac{f(1)}{f(0)},$$which simplifies to $$n(n-1)\cdots (2)(1)=\frac{f(n)}{f(0)}.$$Therefore, $f(n)=n!$ for all $n\ge 1.$

Since $f(9)=9!=326880$ and $f(10)=10!=3628800,$ the smallest such $n$ is $\boxed{10}.$