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:

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