Any way to prove this identity :$\frac{d^m}{dx^m}\frac{H(x)x^{m-1}}{(m-1)!} =\delta(x)$?

One of my friend sent me to proof this identity :$\frac{d^m}{dx^m}\frac{H(x)x^{m-1}}{(m-1)!} =\delta(x)$ Where $\delta$ is the dirac delta function and $H$ is heaviside step function , I knwo only that derivative of the first order of heaviside step function present dirac function, But i never saw the titled formula with $m$ th derivative , Then does the title identity a well known formula ? and how i can show it ?

• what definition of $\delta$ and derivative are you using? – GFauxPas Aug 9 '18 at 0:19
• delta is dirac delta function – zeraoulia rafik Aug 9 '18 at 0:19
• look this :mathworld.wolfram.com/DeltaFunction.html – zeraoulia rafik Aug 9 '18 at 0:21
• Are you sure of this formula ? It's well known that $\frac{d}{dx} H(x) = \delta(0)$, doesn't work for $m=1$ ? – Phoenix Aug 9 '18 at 0:25
• From there, you see the definition of $\delta$ is $\langle \delta, f \rangle = f(0)$, so what are you using to define the derivative of that? – GFauxPas Aug 9 '18 at 0:25