Nth Derivative of x^(n-1) 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.
 A: The following is a proof by induction.

Claim : $\dfrac{d^k}{dx^k} (x^n) = \frac{n!}{(n-k)!}x^{n-k} , 0 \leq k \leq n$

Proof:
The base case is $k=0$, for which the function is $x^n$.
Suppose that we are given that $\dfrac{d^m}{dx^m} (x^n) = \frac{n!}{(n-m)!}x^{n-m}$. Then, 
$$
\dfrac{d^{m+1}}{dx^{m+1}} (x^n) =\dfrac{d}{dx}\dfrac{d^{m}}{dx^{m}} (x^n) =\frac{n!}{(n-m)!}(n-m)x^{n-m-1} =\frac{n!}{(n-m-1)!}x^{n-m-1} 
$$
Hence, the induction is complete.
Now, put $k=n$, and you get $\dfrac{d^n}{dx^n} (x^n) = n!$, which is a constant. Hence, $\dfrac{d^{n+1}}{dx^{n+1}} (x^n) = 0$. Finally, you can replace $n$ with $n-1$ everywhere to complete the proof.
A: Base Case:
$$
(c)'=0
$$
For some constant $c$, a degree 0 polynomial. 
Inductive step: Suppose 
$$
\frac{d^n}{dx^n}x^{n-1}=0
$$
then 
$$
\frac{d^{n+1}}{dx^{n+1}}x^{n}=\frac{d^{n+1}}{dx^{n+1}}x*x^{n-1}\\
=[\frac{d^{n+1}}{dx^{n+1}}x]*x^{n-1}+[\frac{d^{n+1}}{dx^{n+1}}x^{n-1}]*x
$$
by the product rule. But the left term is certainly zero and the right hand side is zero by the inductive hypothesis.
A: Using Leibniz's formula
$(fg)^{(n)}
=\sum_{k=0}^n \binom{n}{k} f^{(k)}g^{(n-k)}
$.
Set
$f(x) = x$.
Since
$f'(x) = 1$
and
$f^{(m)}(x) = 0$
for $m \ge 2$,
$(xg)^{(n)}
=x g^{(n)}(x)+g^{(n-1)}(x)
$.
Now set
$g(x) = x^{n-2}
$.
This becomes
$(x^{n-1})^{(n)}
=x (x^{n-2})^{(n)}(x)+(x^{n-2})^{(n-1)}(x)
$.
If
(this is our
induction hypothesis,
which is true for
$n=2$ and $n=3$)
$(x^{n-2})^{(n-1)}(x)
=0$,
then, also
$(x^{n-2})^{(n)}(x)
=0$
so that
$(x^{n-1})^{(n)}
=0$.
