1A. Find all natural numbers $n$ for which $n^{3}-n=n!$.
Solution. The equation $n^{3}-n=n!$ can be written as $n(n-1)(n+1)=n(n-1)(n-2) \cdots \cdot 3 \cdot 2 \cdot 1$. Since $n=1$ is not a solution to the equation, we will divide both sides by $n(n-1)$. We get $n+1=(n-2)(n-3) \cdots \cdot 3 \cdot 2 \cdot 1$. By direct verification
starting from $n=2$, we find that $n=5$ is a solution. For $n>5, n+1<2n-4=2(n-2)$, that is, $n+1<2(n-2)<(n-2)(n-3) \cdots \cdot 2 \cdot 1$. This means that $n=5$ is the only solution.