Find all positive integers $n$ such that $1!+2!+\ldots+n!$ is a perfect square.
If $n \geqslant 5$, then $1!+2!+\ldots+n!\equiv 1+2+6+24 \equiv 3[5]$. However, modulo 5, we have:

| $x$ | 0 | 1 | 2 | 3 | 4 |
| ---: | ---: | ---: | ---: | ---: | ---: |
| $x^{2}$ | 0 | 1 | 4 | 4 | 1 |

Thus, $n \leqslant 5$. By testing manually, we find that only $n=1$ and $n=3$ work.