Denote $S_n$ as the set of all bijections from $\{1,2,...,n\}\to\{1,2,...,n\}$. Consider the places of the leaves as a bijection $f\in S_n$. Then, from the beginning bijection $g(x) = x,\forall x\in [n]=\{1,2,...,n\}$, each time we clap, we just transform $g\to g\circ f^{-1}$. Hence, we need a bijection $f$ without any fixed point such that $f^{-m}(x)= x,\forall x\in [n]$. By writing $f^{-1}$ in canonical form, we need to express $n = \alpha_1 + \alpha_2 + \cdots + \alpha_k$, such that $\alpha_i$ is a divisor greater than $1$ of $m$. Nevertheless, this is true since we can take two distinct prime divisors $p,q$ of $m$, and let $t = \frac{n}{p}\mod q$, we get $n = tp + sq$ with some $t,s\geq 0$. (Q. E. D)