how can i demonstrate this ? :

$$\ \frac{d^n[x^{n-1}]}{dx^n} =0 \quad, \forall n\in\mathbb N\*$$

i started like this :

$$\ f(x)=x^{n-1} $$

$$\ f'(x)=(n-1)x^{n-2} $$

$$\ f''(x)=(n-2)(n-1)x^{n-3} $$

$$\ f^n(x)= \, ? \, $$ 

maybe something like $$\ (n-1)!x^{(n-1)-n} $$ but $$\ (n-1) < n $$

Thanks a lot.