But There's Infinitely Many Factorials!

For $n=1,2,3,4,5,6$ we check manually .Where we notice for $n=5$ , $n! +5$ is a perfect cube.For $n\geq7$

$n!+5\equiv5\pmod {7}$ .Which is not a cubic residue class modulo 7.

Thus the only positive integer with the desired property is $n=5$ .

The answer is $\boxed{5}$

Let $n! + 5 = y^3$ for some integers $n$ and $y$ . We can explicitly check the cases when $1 \leq n \leq 9$ via a calculator which gives a solution $(n,y) = (5,5)$ .

When $n! \geq 10$ then $25 | n!$ and so $25 \! \not | \> n! + 5$ meaning $25 \! \not | \> y^3$ . Also, since $5 | n! + 5$ , we must have $5 | y^3$ which implies $5 | y$ and $125 | y^3$ . However this causes a contradiction as now $25 | y^3$ . Hence there are no solutions for $n \geq 10$ giving only one of $5$ .

We know that a perfect cube is always of the form $9k + 1$ or $9k - 1$

So , if value of $n > 5$ , then the given no. is of the form $9k + 5$ .

So , testing up the cases till n = 5 yields $\fbox{n = 5}$ as the only solution.

Suppose for some $n \geq 5$ , $n!+5$ is a perfect cube. Then $n!+5=5{(n-1)!+1}$ So $(n-1)!+1$ must have a unit digit 5. Or $(n-1)!$ must end in 4. This can't happen for $n<4$ Thus, $\boxed {n=5}$ is the only solution.