For a geometric sequence {a_n}:

a_n = a_1*r^{n - 1}

where a_1 is the first term in the series and a_n is the nth term.

So

a_5 = 5! = 120 = a_1*r^4

a_6 = 6! = 720 = a_1*r^5

Two equations, two unknowns, we can solve this.

Divide a_6 by a_5:

a_6/a_5 = 6!/5! = 6 = (a_1*r^5)/(a_1*r^4) = r

Thus r = 6.

So

120 = a_1*6^4

Thus

a_1 = 120/6^4 = 20/6^3 = 20/216 = 5/54

So

a_n = (5/54)*6^{n - 1}

So

a_4 = (5/54)*6^{4 - 1} = (5/54)*6^3 = (5/54)*216 = 20

-Dan