The function f(t) = t^n is continuous on [0,x], for any x > 0 ==> for any value c in (f(0),f(x)) = (0, x^n) there exists some d in (0,x) s.t. f(d) = c ==> for any positive c there exists an expression d^n = c for it (since any c is contained in some interval of the fomr [0,x] for some positive x, because f is increasing monotonically ==> c^(1/n) = d and c has a positive n-th root d.

About uniqueness it follows at once from (a).

Tonio