A number theory problem by tanveen dhingra

If $n \ge 4$ then $n!$ is a multiple of $8$ , so that $n! +6$ is $2$ times an odd number. Thus the index of $2$ in $n!+6$ is $1$ , and so $n!+6$ cannot be a cube (or, for that matter, any power). In the range $1 \le n \le 3$ , only $n=2$ gives a cube ( and no value of $n$ gives a square or any other power).

The cubic residues modulo 4 are $0,1,3$ .When $n\geq 4$ then $n!+6\equiv 2\pmod{4}$ which is not a cubic residue modulo 4.Hence $n<4$ .Simple case checking reveals that $n=\boxed{2}$ gives the only solution.

Modular Arithmetic is the way to go! $$ We first consider the cubic residues $\pmod{8}$ ,and we find that no cube leaves a remainder of $6$ when divided by 8,hence we get that $n<8$ ,when we consider the different values of $n$ we get that only $n=2$ satisfies the conditions.And we are done!