No.

I believe the BS you've fallen for / trying to propagate is to prove

n^2 = n(2n-1) for all n because

(a) 1^2 = 1(2*1-1) --> OK

(b) if x^2 = x(2x-1) then (x+1)^2 = (x+1)(2x+1)

LHS = (x+1)^2 = x^2 + 2x + 1 = x(2x-1) + 2x + 1 = 2x^2 + x + 1

Not sure how you equate it to RHS

saying

(x+1)(2*(x+1)-1) = (x+1)^2

Proof: (x+1)(2*(x+1)-1) = (x+1)^2 QED

is restating the same line. Not a proof.