The expression (x+y)ⁿ is

(x+y)(x+y)…(x+y)
|←------n times----→|

This expression evaluates to the sum of terms involving x^ky^n-k for k = 0 to n. The number of times the term x^ky^n-k comes up is the number of ways of choosing x k times from the n binomials. Let that number be denoted as (ⁿ_k). Therefore

(x+y)ⁿ = Σ_k=0ⁿ (ⁿ_k)x^ky^n-k

Q.E.D.

The Evaluation of (ⁿ_k)
the number of times x^ky^n-k occurs

The first x can be chosen from n different places. The second for (n-1) different places. The third from (n-2) different places. Thus the k-th x can be chosen from (n-(k-1)) different places. The total number is thus given by n(n-1)·…·(n-(k-1)). By the definition of the factorial function this is the same as n!/(n-k)!. It does not matter among the k choices of the places of x which one is first or second or so forth. This means the k! arrangements of the choices of x's are the same. Therefore the number of distinct ways of choosing x k times from the n binomials is

(ⁿ_k) = n!/(k!(n-k)!)

Corollaries

The most obvious corollary is that

(ⁿ_n-k) = (ⁿ_k)

Note that if x=1 and y=1 then the Binomial Theorem proves that

Σ_k=0ⁿ (ⁿ_k) = 2ⁿ

Likewise if x=−1 and y=1 then the Binomial Theorem proves that

Σ_k=0ⁿ (ⁿ_k)(−1)^k = 0

Reference:

Arthur Benjamin and Jennifer J. Quinn, Proofs that Really Count: The Art of Combinatorial Proof, Mathematical Association of America, 2003.