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, \[g(n) g(n - 1) \dotsm g(2) g(1) = \frac{f(n)}{f(n - 1)} \cdot \frac{f(n - 1)}{f(n - 2)} \dotsm \frac{f(2)}{f(1)} \cdot \frac{f(1)}{f(0)},\]which simplifies to \[n(n - 1) \dotsm (2)(1) = \frac{f(n)}{f(0)}.\]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} $.