Yes, but similar flaws exist for your proof.

The algebraic proof that 0.999… = 1 must first prove why you can assign 0.999… to x.

My “proof” abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.

The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999… will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999… is 1.

Unfortunately not an ideal proof.

It makes certain assumptions:

1. That a number 0.999… exists and is well-defined
2. That multiplication and subtraction for this number work as expected

Similarly, I could prove that the number which consists of infinite 9’s to the left of the decimal separator is equal to -1:

...999.0 = x
...990.0 = 10x

Calculate x - 10x:

x - 10x = ...999.0 - ...990.0
-9x = 9
x = -1

And while this is true for 10-adic numbers, it is certainly not true for the real numbers.

Imagine if the algorithm were in Θ(n!²), that would be even worse