A number theory problem by mostafa hubble

Consider $n \geq 6$ Then $n!$ can be written as $3^{k}* q$ with $k\geq2$ . Then $3^{n-1} = n! + 3 = 3^{k}*q + 3 = 3*(3^{k-1}q +1)$ which is not a perfect power of 3 since the second factor is not a multiple of 3. Taking modulo 3 on the first eqution we see $n!$ must be divisible by 3. That means $5 \leq n \leq 3$ Checking cases we see $n = 3$ and $n = 4$ work. So there are 2 options.

Since the factorial function grows much faster than the exponential function, we just have to try with the first 4 natural numbers. Taking into account that 5! is bigger than 3^4.