I am stuck on the end part here (or maybe im not) with this proof

Prove n^2-1 is a multiple of 8 whenever n is an odd integer.

Heres what I have done

Let n = even integer

therefore n = 2K k integer (im still unfamiliar with getting the symbols on the forums)

(2k)^2 - 1 is a multiple of 8?

4k^2 - 1 is multiple of 8,

4k^2 = even. even - 1= odd so that doesnt work

so

Let n = odd integer

Therefor n = 2k + 1 k integer

(2k+1)^2 - 1 multiple of 8

4k^2 + 4k +1 - 1 is a multiple of 8

4k^2 + 4K is a multiple of 8.

4k(k+1) is a multiple of 8..

(k+1) is an integer.

4k(k+1) is a multiple of 8. QED

Is this correct, or am I missing any parts?