1. Find all natural numbers $n$ such that the number

$$
1!+2!+3!+\ldots+n!
$$

is a perfect square. (Here: $k!=1 \cdot 2 \cdot \ldots \cdot k, 1!=1$)
Solution. From the fact that $5!=120$, we conclude that $n!$ ends in zero for every $n \geq 5$. According to this, the number $1!+2!+3!+\ldots+n!$ ends in 3 for $n \geq 5$, (since the number $1!+2!+3!+4!=33$ ends in 3). If a number ends in 3, then that number is not a square of any number, because the square of any number ends in 1, 4, 5, 6, 9, or 0. Therefore, for no natural number $n \geq 5$ is the number $1!+2!+3!+\ldots+n!$ a perfect square. It remains to check whether for some natural number $n \leq 4$ the number $1!+2!+3!+\ldots+n!$ is a perfect square. Specifically:

$$
\begin{array}{ll}
\text { - for } n=1, & 1!+2!+3!+\ldots+n!=1=1^{2} \\
\text { - for } n=2, & 1!+2!+3!+\ldots+n!=1!+2!=3, \\
\text { - for } n=3, & 1!+2!+3!+\ldots+n!=1!+2!+3!=9=3^{2} \\
\text { - for } n=4, & 1!+2!+3!+\ldots+n!=1!+2!+3!+4!=33
\end{array}
$$

Thus, the only natural numbers $n$ for which the number $1!+2!+3!+\ldots+n!$ is a perfect square of some natural number are 1 and 3.