This question was discussed many many times in this forum:
In order to better understand this problem , I suggest to plug in some values.

h(n) where n is positive and even (e.g 2,4,6) and h(n) is the prouduct of all even numbers from 2 to n.

n=10

h(10) = 10*8*6*4*2

so after we understand what this problem wants , we can now solve it !

h(100)+1 = (100*98*96*94....6*4*2) + 1 can be simplified to (2^50)*50!+1

what is the smallest prime factor ?

For any factorial + 1 the smallest factor (apart from 1) is greater then any of the members of the factorial

2! + 1 = 3: smallest factor is 3

3! + 1 = 7: smallest factor is 7

4! + 1 = 25: smallest factor is 5

5! + 1 = 121: smallest factor is 11

6! + 1 = 721: smallest factor is 103

50! + 1 = a number greater then 50

hope this will help.