Find the value of the Euler phi-function at each of these integers

Please help me with these problems!
a). 100
b). 256
c). 1001
d). 2*3*5*7*11*13
e). 10!
f). 20!

Use prime factorizations along with φ(p^n) = p^n - p^(n-1).

a) φ(100) = φ(2^2 * 5^2) = φ(2^2) * φ(5^2) = (4 - 2)(25 - 5) = 40.

b) φ(256) = φ(2^8) = 2^8 - 2^7 = 128.

c) φ(1001) = φ(7 * 11 * 13) = φ(7) φ(11) φ(13) = (7 - 1)(11 - 1)(13 - 1) = 7200

d) As with c, φ(2*3*5*7*11*13) = 1 * 2 * 4 * 6 * 10 * 12 = 5760.

e) φ(10!) = φ(2^8 * 3^4 * 5^2 * 7) = (2^8 - 2^7) (3^4 - 3^3) (5^2 - 5)(7 - 1) = 829440

f) Similar to part e:
φ(20!) = φ(2^18 * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19)
.........= (2^18 - 2^17) (3^8 - 3^7) (5^4 - 5^3) (7^2 - 7) (11 - 1) (13 - 1) (17 - 1)(19 - 1).

I hope this helps!

The Euler Phi function φ(x) can be simplified to be the product of (p^k - p^(k-1)) for all the prime factors p of x, with exponent k. So:

a) 100 = 2^2 * 5^2, so φ(100) = (4-2)* (25-5) = 2*20 = 40
b) 256 = 2^8, so φ(256) = 2^8 - 2^7 = 128
c) 1001 = 11 * 13 * 17, so φ(1001) = 10*12*16
d) φ(2*3*5*7*11*13) = 1*2*4*6*10*12
e) 10! = 2^8 * 3^4 * 5^2 * 7, so φ(10!) = (2^8-2^7)(3^4-3^3)(5^2-5)(6)
f) 20! = 2^18 * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19
φ(20!) = (2^18-2^17) * (3^8-3^7) * (5^4-5^3) * (7^2 - 7) * 10 * 12 * 16 * 18