Looks simple, doesn't it?

I can build on Finn's partial solution. As he notes, the sum can't be a square unless $n = 3$ . We can check one by one that for $n \le 8$ the sum is not a $b$ th power. Now for $n = 8$ the sum is $46233$ , which is divisible by $9$ and not $27$ . Since $k!$ is divisible by $27$ for $k \ge 9$ , the sum will be divisible by $9$ and not $27$ for $n \ge 9$ . So it can't be a $b$ th power for $b \ge 3$ either.

So the only solution is $n = 3$ , which leads to $\fbox{64}$ .

I have an immensely powerful tool for approaching this problem. It's easy to find the first solution, which is $(3, 3, 2)$ , but how to prove it is the only and the largest?

It should not be too hard to figure out all the different sums.

At $n=2$ , the sum is $3$ . Similarly, letting $f(n)$ be the summation, we get

$f(3)=\boxed{9}$ ,

$f(4)=33$ ,

and $f(5)=153$ .

Now notice something about factorials. Everything after $4!$ ends with a zero, like $120$ , $720$ , etc. Consider all squares $\mod{10}$ . They are:

$1$ ,

$4$ ,

$9$ ,

$6$ ,

and $5$ .

These are the only possible remainders. Why am I saying all this? Because look at $f(4)$ and $f(5)$ . They both end in $3$ . Because, as we've pointed out, by adding the next highest factorials, the units digit will not be affected. From what we've collected about squares modulo $10$ , it is impossible for any $f(n)$ for $n \geq 4$ to be a perfect square. As for cubes and any higher powers, I am clueless. Otherwise, we have shown that $n=3$ is the only solution and thus the required expression is $(3+3+2)^2=8^2=\boxed{64}$ . Absolutely fantastic problem @Victor Loh but I have no idea how to prove this for higher powers. :O