Ah, but infinity is not ONLY inclusive of 0 (as in, there are an infinite number of integers) - but it is SIMULTANEOUSLY inclusive of zero and EVERY OTHER number AT THE SAME TIME.
Since infinity is not a scalar number, but rather is an ambiguous concept that has a (literally infinite) number of possible members in a set, use of the mathematics operators is not technically valid. Your proof fails on the violation of using the math operators on a number that is outside of their intended scope.
You can do anything mathematical you want to infinity and it is still infinity. The math operators have NO EFFECT on it. You need special cases for those situations where you CAN affect infinity. For instance, there are multiplicative situations where you can reach aleph-1 (where aleph-0 is the ordinary, run-of-the-mill infinity), aleph-2, etc. There are "higher levels" or perhaps "higher orders" of infinities.
When it comes to the standard math operators, infinity is generally outside of their scope of influence. So for THAT reason, "infinity - 1 = infinity." BUT your proof fails because the math symbols have no effect on infinity - but they DO take effect on ordinary numbers. So in USA football, the yellow penalty flag would come out for "Illegal shift" or "Illegal formation."
You question about "recurring" progressions is in alignment with limit theorems where a progression is taken to its extreme. (Egad! It has been literally FIFTY YEARS since I dealt with this stuff...). If you take this progression (and sorry, but I'm not going graphics on this
X = Sum (for I ranging from 1 to infinity) of ( 9 divided by (ten to the Ith power) )
The first few steps are 0.9, 0.99, 0.999, 0.9999, (that's the first four, you have an infinite number of steps remaining). The LIMIT of that series is 1, though mathematically, it will never actually reach that value. Another variant of Zeno's paradox, more or less.