What is the value of (n!)^1/n where n tends to the infinity?

Question

If 1^3 + 2^3 + 3^3 = m^2 where m is also an integer.
What are the next three consecutive positive integers such that the sum of their individual cubes is equal to a perfect square?

Answer 1

n! tends to infinity and 1/n tends to zero, the resulting value is going to be 1 as zero power of any number.