Let's determine all natural numbers for which $1!+2!+\ldots+n!$ is a perfect square (where $n!$ denotes the product of the first $n$ natural numbers).
To solve the task, it is enough to look at the last digit of the questioned sum. If $n \geq 5$, then $n!$ (read: $n$ factorial) ends in 0, meaning it is divisible by 10, since 2 and 5 appear as prime factors. Therefore, if $n \geq 4$, the sum $1!+2!+\ldots+n!$ ends in the same digit as $1!+2!+3!+4!=33$, which is 3. However, there is no square number that ends in 3, since the square of a number's last digit cannot end in 3 $\left(1^{2}=1,2^{2}=\mathbf{4}, 3^{2}=\mathbf{9}, 4^{2}=16,5^{2}=25,6^{2}=36,7^{2}=49,8^{2}=64,9^{2}=81,0^{2}=\mathbf{0}\right)$, none of these last digits is 3. Thus, the expression $1!+2!+\ldots+n!$ for $n \geq 4$ cannot be a perfect square.

Let's examine the sum for $n<4$:

$n=1, \quad 1!=1=1^{2}$, this is a square number,

$n=2, \quad 1!+2!=3$, not a square number,

$n=3, \quad 1!+2!+3!=9=3^{2}$, this is also a square number.

Therefore, the sum is a perfect square for $n=1$ and $n=3$.