n=1 works out to 1-1+4 = 4 which has a factor of 2.
Assume n4-n+4 is even for n=k (and that expression in terms of k is simply k4-k+4)
Now for n=k+1 we get (k+1)4-(k+1)+4 which needs to be shown to have a factor of 2.
(k+1)4 -(k+1) + 4 = (k2+2k+1)2-(k+1) + 4
= (k4 + 4k3 + 6k2 + 4k + 1) - (k+1) + 4
= k4 + 4k3 + 6k2 + 3k + 4
Now look at the difference between k4 -k+4 and this last expression:
(k4+4k3+6k2+3k+4) - (k4-k+4)
This last expression has a factor of 2 (i.e. it is even).
So we know k4-k+4 has a factor of 2 (by induction hypothesis), and if we add the even number 2(2k3+3k2+2k) to it, we get the value for n=k+1. Therefore n4-n+4 always has a factor of 2.