Well, yeah, the denominator is n! isn't it!

compare the numerators to the power of x: when

k= 1 numerator is n

k= 2 numerator is n(n-1)

k= 3 numerator is n(n-1)(n-2)

k= 4 numerator is n(n-1)(n-3)

Isn't it obvious that this is n(n-1)(n-2)...(n-(k-1))?

If you are thinking you need to write that as a "closed form", consider that

n(n-1)(n- 2)... (n-a)= [n(n-1)(n-2)...(n-a)(n-a-1)...(3)(2)(1)]/(n-a-1)(n-a-2)...(3)(2)(1)]= n!/(n-a-1)!