numbers and factors 2

well its easy
See for n ≥ 10
last two digits will be 00 but
as any cube of form (5k) ends in 125
so n!+5 will end in 05 for n≥10

Jut check for n<10
answer is n=5 :)

Alternative: (A.Titu)

for , n\geq 7 , n! + 5 \equiv 5 (mod 7) a contradiction for all cubes. we check for only n< 7 [6]

Well ya a great solution indeed :)