Fermat numbers question

F subk = 2^(2^k) +1 where k is greater than or equal to 0.

calculate f sub 0, f sub 1, f sub2, f sub3, f sub4 and the product fsub3 x f sub2 x f sub1 x f sub0 and verify that f sub4 -2= f sub3 x f sub2 x f sub1 x f sub0

F[0] = 2^(2^0) + 1 = 2^1 + 1 = 2 + 1 = 3
F[1] = 2^(2^1) + 1 = 2^2 + 1 = 4 + 1 = 5
F[2] = 2^(2^2) + 1 = 2^4 + 1 = 16 + 1 = 17
F[3] = 2^(2^3) + 1 = 2^8 + 1 = 256 + 1 = 257
F[4] = 2^(2^4) + 1 = 2^16 + 1 = 65536 + 1 = 65537

You can do the rest.