From this calculation
f[x_] := x^n - x^(n - 1);
f'[x] // Factor
sol = SolveValues[f'[x] == 0, x]
f''[sol[[1]]] // Factor
we see that (n-1)/n is always a local minimum, because the first derivative vanishes and the second derivative is positive.
The SolveValues above misses x==0, which is also a stationary point, but from
In f[x] // Factor
Out (-1 + x) x^(-1 + n)
we see that x==0 is either a local maximum or a flection point.
