arithmetic - False proof $m = 1$ for any $m$

Is it possible to find a fake proof that results in $$ 1 = m, $$ where $m$ is any number you chose?

The idea is to find a generalization to this fallacy:

\begin{align}a &= b\\
&\Downarrow\\
a^2 - b^2 &= ab - b^2\\
(a+b) (a-b) &= b (a-b)\\
a+b &= b\\
2b &= b \\
2 &= 1 \end{align}

You can use division by zero or anything a little bit elaborated you can think of.

Answer

Using $2=1$, you can prove using induction that $$\forall n\in\mathbb N,\,n=1.$$

Indeed, it is true for $1$ and also for $2$ by assumption. Assume that it's true for a certain $n>2$.

\begin{align}
n+1&=n-1+2\\
&=n-1+1\\
&=n\\
&=1.

\end{align}

Edit:

Another method which looks like a real generalisation of your given proof as follows: let $n\in\mathbb N$. We have:

\begin{align}
a&=b\\
a^n-b^n&=ab^{n-1}-b^n\\
(a-b)\sum_{k=0}^{n-1}a^kb^{n-1-k}&=b^{n-1}(a-b)\\

\sum_{k=0}^{n-1}a^kb^{n-1-k}&=b^{n-1}\\
nb^{n-1}&=b^{n-1}\\
n&=1.
\end{align}