If you want to include more than one solution:
e^(i x pi x (1 + 2Z)) + 1 = 0
where (1 + 2Z) is shorthand for the odd integers. So your equation is simply shorthand for two of the above, and not really anymore satysfying. In fact, I much prefer
e^(i x pi) + 1 = 0
because there's no amiguity, and the symbols are all used exactly one time. Also remember that e^(i x pi) is -1, so it's plus, not minus.
Slashdot Mirror
If you want to include more than one solution: e^(i x pi x (1 + 2Z)) + 1 = 0 where (1 + 2Z) is shorthand for the odd integers. So your equation is simply shorthand for two of the above, and not really anymore satysfying. In fact, I much prefer e^(i x pi) + 1 = 0 because there's no amiguity, and the symbols are all used exactly one time. Also remember that e^(i x pi) is -1, so it's plus, not minus.