# Correctly restricting domain of parameters within Solve

I have seen a few posts about this (such as this one), but none solved my problem. As a silly example, suppose I want to solve (1-x)^n = 0, knowing that n > 1. I tried doing,

Assuming[n > 1, Solve[(1 - x)^n == 0, x]]


But, Mathematica goes a bit crazy and gives,

{x -> 1 - 0^(1/n)}


with a warning message. Now, the answer is technically correct I guess, but I would really just like to see x -> 1 of course. Note that Mathematica probably ignores the assumption, because, if I assume n < -1, then I get the same output anyway, although there really is no solution. Is there a way to make this work as expected?

• How about Solve[(1 - x)^n == 0 && n > 1, x] ? – Nasser Jul 30 '19 at 12:56
• Ok, that works, although I fail to see what was wrong with my attempt? Thanks anyway. – Patrick.B Jul 30 '19 at 13:00