Finding some explicit formula for $(ab)^n$ in any $a,b$ in a finite $p$-group. 
If $G$ is a finite $p-$group, let $a$ and $b$ any two elements from $G$.
Is there any formula for $(ab)^n$ involving $a^nb^n$ for any natural number $n$? That is, some formula like $(ab)^n = a^nb^nf(a,b)$ for some function of $a$ and $b$.

Please not those "modulo" some subgroups formulas! Like $$
(ab)^n \equiv a^nb^n \pmod{H},
$$
where $H \leq G$
Thanks in advance.
 A: There are some formulas* and they involve (often unknown) commutators. Sometimes there are particularly nice formulas, but they hold in $p$-groups and for things of the form $(ab)^{p^n}$, so they do not hold for arbitrary powers.
Some example formuals are:


*

*If $G$ is nilpotent of class at most two then the identity $$(xy)^m=x^my^m[y,x]^{m\choose2}$$ holds. Which is really pretty. You can generalise this, via the Hall-Petresco formula, to groups of higher class, but the cost here is more commutators.

*If $G$ is any group then we have the Hall-Petresco formula, as mentioned in the comments: $$x^my^m=(xy)^mc_2^{m\choose {2}}c_3^{m\choose {3}}\cdots c_{m-1}^{m\choose {m-1}}c_m.$$ Here each $c_i$ is contained in the $i^{th}$ subgroup of the decending central series for the group $G$. See Section 12.3 of M. Hall (1959), The theory of groups, Macmillan, MR 0103215. This book also contains some useful formulas for regular $p$-groups. Note that it is Philip Hall after whome the formula is named, while Marshall Hall wrote a book (and did other stuff too!).
You can find other formulas in the first section of the book C. Leedham-Green, S. McKay (2002) The Structure of Groups of Prime Power Order, Oxford University Press, and also in the book Y. Berkovich (2008) Groups of Prime Power Order Volume 1 (Appendix 1 of this book proves the Hall-Petresco formula).
*or formulae, if you want to be correct but also sound slightly pretentious :-)
