We can write an = (a–b+b)n
Then an = [(a−b)+b]n
=

n

‌

C0(a−b)n+

n

‌

C1(a−b)n−1b + ... +

n

‌

Cn−1(a−b)bn−1 +

n

‌

Cnbn
⇒ an−bn =

n

‌

C0(a−b)n+

n

‌

C1(a−b)n−1b + ... +

n

‌

Cn−1(a−b)bn−1+bn−bn
= (a - b) [

n

‌

C0(a−b)n−1+

n

‌

C1(a−b)n−2b + ... +

n

‌

Cn−1bn−1]
= (a – b) (some integer)
[

n

‌

C0,

n

‌

C1,

n

‌

C2 , ... ,

n

‌

Cn−1 , are integers & also all non negative powers of a – b and b are integers]
Hence, a – b is a factor of an–bn.