Compute for $m \in \mathbb N \setminus \{0\}$, $ \ x \in \mathbb R$, $$ \frac{d^m}{dx^m} \Big( H(x) \frac {x^{m-1}}{(m-1)!}\Big) \ , $$ where $H(x)$ is the Heaviside function.
My solution:
For the general Leibniz rule, \begin{align} \frac{d^m}{dx^m} \Big( H(x) \frac {x^{m-1}}{(m-1)!}\Big) &= \sum_{k=0}^m \binom {m}{k} \frac{d^{m-k}}{dx^{m-k}} H(x) \frac{d^k}{dx^k} \frac {x^{m-1}}{(m-1)!}\\ &= \sum_{k=0}^{m-2} \binom {m}{k} \frac{d^{m-k}}{dx^{m-k}} H(x) \frac{d^k}{dx^k} \frac {x^{m-1}}{(m-1)!}\\& + \binom{m}{m-1} \frac{d}{dx}H(x) \frac{d^{m-1}}{dx^{m-1}} \frac {x^{m-1}}{(m-1)!}\\ &+ \binom{m}{m}H(x) \frac{d^{m}}{dx^{m}} \frac {x^{m-1}}{(m-1)!} \end{align} but $\frac{d}{dx} H(x) = \delta (x)$, so $$ =\frac{1}{(m-1)!} \Big( \ \sum_{k=0}^{m-2} \binom {m}{k} \frac{d^{m-k-1}}{dx^{m-k-1}} \delta(x) \frac{d^k}{dx^k} x^{m-1} \ \Big) \ + \ m \, \delta(x) \ \ . $$ I know that if $a(x) \in C^\infty(\mathbb R) $ and $u \in \mathcal D'(\mathbb R)$ then $ a \, u(\varphi) = u (a \, \varphi) \ \ \forall \ \varphi \in C_c^\infty(\mathbb R)$. Therefore, $$ \frac{d^{m-k-1}}{dx^{m-k-1}} \delta(\varphi) \frac{d^k}{dx^k} x^{m-1}= \frac{d^{m-k-1}}{dx^{m-k-1}} \delta(\varphi \frac{d^k}{dx^k} x^{m-1} ) $$ but $k=0,...,m-2\ $, so $$ \frac{d^k}{dx^k} x^{m-1}= \frac{(m-1)!}{(m-k-1)!} x^{m-k-1} \ \ . $$ I deduce that $$ \delta(\varphi \frac{d^k}{dx^k} x^{m-1} )= \delta(\varphi \frac{(m-1)!}{(m-k-1)!} x^{m-k-1})= 0 $$ $ \forall \ \varphi \in C^\infty_c(\mathbb R) \ $ and $\ \forall \ k=0,...,m-2$.
Finally, to conclude, $$ \frac{d^m}{dx^m} \Big( H(x) \frac {x^{m-1}}{(m-1)!}\Big) = m \, \delta(x) \ \ . $$ Is this reasoning correct?