Can you prove (0^0 = 1) -> (1 = 2), as you claimed above?
And no, we don't need “temporary definitions”. For any combinatorial investigation imaginable leads unambiguously to 0^0 = 1:
https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer...
The only concept a general expression of the form x^y could possibly represent is the space of mappings y -> x. There's one and only one such mapping when x and y represent empty set. Arguing this to be wrong as in “but it can't be 1 because there are no mappings to empty space from non-empty sets!” ( = “0^x = 0”) is plain ridiculous. This definition simply does not fall apart. 0^0 is not a special case for it in any way.
Enumerating numbers one can represent with 0 digits put in a string of length 0 leads to the same conclusion. It's clear that there is one and only one string of length 0, unless you demand it to contain more than 0 digits, in which case there are none. [However, this combinatorial problem is not an independent one: it's equivalent to enumerating mappings from space of strings to space of chars.]
In other words, not only there are definitions that work well for all cases, including 0^0, there's actually only one such definition.