Geometric Sum proof

So I was doing a past paper and had the "Prove the sum of a geometric series is ..."

I constructed a proof that in my opinion seems to make sense, but wasn't recognised on the mark scheme so just wondered whether it was actually valid. To be honest, I'll probably just use the standard one in future, but am curious.

Sn = a + ar+ar^2+ar^3+..+ar^n-1
Sn = a(1 + r^2+...+r^n-1)
Sn = (a(1 + r^2+...+r^n-1)(1-r)) / (1-r)
Sn = (a(1 - r+r-r^2+r^2-r^3+r^3...r^n-1+r^n-1-r^n)) / (1-r)
Sn = (a(r - r^n)) / (1-r)

Seems ok. The factorisation
(1 - x^n) = (1-x)(1+x+...+x^(n-1))
is fairly well known (if youre doing further, then you sometimes use it in complex roots of unity) and your way does the old algebraic trick of multiply by 1 = (1-r)/(1-r). Though the end result should be a(1-r^n) on the numerator?

If you can assume the infinite sum, taking the difference of two infinite sums though is almost 1 line proof and personally I like the recursive definitions/proofs, but theyre all fairly similar.

Brill, thanks. Messed up the last line on here!

Sorry to be a pain, but would you mind expanding on what you mean about taking the difference of 2 infinite sums?

Assuming you have
S1..inf = a/(1-r)
Then
S1..n = S1..inf - S(n+1)..inf = a/(1-r) - ar^n/(1-r) = a(1-r^n)/(1-r)

It just uses the basic series fact that
S1..inf = S1..n + S(n+1)..inf