So here is how you do a proof by weak induction although in this case it is using a nuclear weapon to swat a fly.
What is to be proved must be stated as a proposition about an integer as in
[math]\text {For any positive integer } n \ge 1 \text { and real } x \ne 1, \\
1 - x - x^n + x^{(n+1)} = (1 - 2x + x^2) * P_n(x),\\
\text {where } P_n(x) \text { is a polynomial of degree } n - 1.
[/math]
So we must prove it for n = 1 first. But that is trivial.
[math]
n = 1 \implies 1 - x - x^n + x^{(n+1)} = 1 - x - x - x^2 = 1 - 2x - x^2 = (1 - 2x + x^2) * 1, \text { and}\\
1 \text { is a polynomial of degree } 0 = 1-1.
[/math]
So our proposition is certainly true for one or more integers.
[math]
\text {Let } k \text { be an arbitrary integer such that } k \ge 1 \text { and}\\
1 - x - x^k + x^{(k+1)} = (1 - 2x + x^2) * P_k(x), \text { where}\\
P_k(x) \text { is a polynomial of degree } k - 1.\\
\text {But } y = 1 - x - x^{(k+1)} + x^{\{(k+1)+1\}} = 1 - x + 0 + 0 - x^{(k+1)} + x^{\{(k+1)+1\}} \implies\\
y = 1 - x + (x^k - x^k) + (2x^{(k+1)} - 2x^{(k+1)}) - x^{(k +1)} + x^{\{(k+1)+1\}} \implies \\
y = (1 - x - x^k + 2x^{(k+1)} - x^{(k+1)} ) + (x^k - 2x^{(k+1)} + x^{\{(k+1)+1\}} \implies \\
y = (1 - x - x^2 + x^{(k+1)}) + (1 - 2x + x^2) * x^k) \implies \\
y = (1 - 2x + x^2) * P_k(x) + (1 - 2x + x^2) * x^k = (1 - 2x + x^2) * \{x^k + P_k(x)\}.\\
\text {Let } P_{k+1}(x) = x^k + P_k(x).\\
\therefore P_{k+1} \text { is a polynomial of degree } k = (k+1)-1\\
\because \ P_k \text { is a polynomial of degree } k - 1.\\
\therefore \ 1 - x - x^{(k+1)} + x^{\{(k+1)+1} = (1 - 2x + x^2) * P_{k+1}(x).\\
\text {THUS, the proposition is true for any integer } n \ge 1.
[/math]
Such proofs are very important, but you do not need this technique for this proposition. What is a simpler proof?