when is $n!+10$ a perfect square?
Solution 1:
In base $10$, the digits of a perfect square must end in $1, 4, 6, 9, 00$, or $25$. However, for $n\geq 10$, $n!$ is divisible by $100$ and so the last two digits of $n!+10$ are $10$. Therefore, $n!+10$ cannot be a perfect square for $n\geq 10$. And you can just check directly that for $1\leq n\leq 10$, only $n=3$ gives a perfect square.
Solution 2:
A perfect square is congruent to $0$ or $1 \bmod 4$.
If $n\ge 4$ then $n! \equiv 0 \bmod 4$ and so $n!+10 \equiv 2 \bmod 4$ is not a square.
The case $n \le 3$ is dealt by hand.