There is well-known Leibniz rule generalization for the $n$-th derivative of product of $m$ functions $f_1, f_2, \ldots, f_m$, namely: $$ D^n(f_1 f_2 \cdots f_m)=\sum_{k_1+k_2+\cdots+k_m=n} \binom{n}{k_1 \,k_2 \, \cdots k_m} D^{k_1}(f_1)D^{k_2}(f_2)\cdots D^{k_m}(f_m). $$ Is there any simplification of the formula for the case $f_1=f_2=\cdots=f_m=f$?
I hope there is a formula without the multinomial coefficients.
You can use this simplified version of Di Bruno formula:
$${d^n \over dx^n} g(f(x)) = \sum_{k=1}^n g^{(k)}(f(x))\cdot B_{n,k}\left(f'(x),f''(x),\dots,f^{(n-k+1)}(x)\right),$$ which in the case $g=x^m$ gives $${d^n \over dx^n} f^m(x) = \sum_{k=1}^n\frac {m!}{(m-k)!} f^{m-k}(x)\cdot B_{n,k}\left(f'(x),f''(x),\dots,f^{(n-k+1)}(x)\right).$$
But be aware that the multinomial coefficients did nor disappear completely. They are inside the incomplete Bell polynomials $B_{n,k}$.