Earlier quoted context omitted.
Even in non-standard models of arithmetic, where you could use infinity as an operand, the value of that expression would have to be undefined (not 1, as you seem to expect).
What about limit as x goes to infinity of x - (x - 1)? We do the algebra first, right? Now, that's a special case of y - (x - 1) where y = x, so can we get a different residue by going about the limit in two dimensions?
Also, http://www.wolframalpha.com/input/?i=infinity+-+(infinity+-+...
If nothing else, Wolfram Alpha is useful for math. I'm impressed that it understood my syntax for limits that I made up on the spot, first try.