NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1
i.e.
n! + prime(n) != m^k


I assume "integral" really meant integer since no differential is given.

n=3

then

prime(3) = 2

substituting:

n! + prime(n)! = 3! + 2! = 8

2^3 = 8

Therefo

n! +prime(n)! = m^k

where n = 3, m =2 and k = 3

All integers and k1.