Calculating the index of a subgroup in a free group Can someone help me with the following question?

Let $F=\langle x_1,\ldots,x_n \rangle $ be a finitely-generated free group. Let $p$ be a fixed prime, and $H$ be the normal subgroup of $F$ defined by: $H= F^p [F,F]$ .
Prove that $|F:H| = p^n $.

I hope someone will help me!
Thanks!
 A: I have an idea: let us check the $\,n\,$ elements
$$x_1H\,,\,x_2H,\ldots,x_nH\in F/H$$
and suppose we have integers $\,e_i\in\{0,1,...,p-1\}\,\,,\,i=1,2,...,n\,$ ,  s.t.
$$\prod_{k=1}^nx_k^{e_k}H=H\Longleftrightarrow\prod_{k=1}^nx_k^{e_k}=xg\in H\,\,,\,x\in F^p\,\,,\,g\in F'\Longrightarrow$$
$$\Longrightarrow x^{-1}\prod_{k=1}^nx_k^{e_{k}}\in F'$$
We can put $\,x=\prod_{i=1}^rw_i^p\,$ , where the $\,w_i\,$ are (free) words in the letters $\,x_1,...,x_n\,$ . Now a definition:
Definition: If $\,w\in F\,$ , then the exponential sum $\,e_{x_i}\,$ of $\,x_i\,$ in $\,w\,$ is the sum of all the exponents of $\,x_i\,$ in $\,w\,$.
For example, if $\,w=x_1^3x_2x_4^{-3}x_1^{-2}x_6x_1\,$ ,then $\,e_{x_1}=3-2+1=2\,\,,\,e_{x_2}=1\,\,,\,e_{x_4}=-3\,$ , etc.
Thus, the exponential sum of all the generators $\,x_1,...,x_n\,$ in $\,x\,$ is a multiple of $\,p\,$, since $\,x\,$ is the product of some words $\,w_1,...,w_s\in F\,$ to the $\,p-$th power, and the same is true for $\,x^{-1}\,$ , of course.
Then, the exponential sum of each $\,x_i\,$ in $\,x^{-1}\prod_{k=1}^nx_k^{e_k}\,$ is of the form $\,t_ip+e_i\,\,,\,\,t_i\in\Bbb Z\,$ .
But since a word $\,w(x_1,...,x_r)\,$ in free generators is a commutator element iff every word's letter's exponential sum is zero, we get
$$t_ip+e_i=0\Longleftrightarrow t_i=e_i=0$$
since $\,0\leq e_i\leq p-1\,$ .
This proves the $\,n\,$ elements above are linearly independent over $\,\Bbb F_p:=\Bbb Z/p\Bbb Z\,$ and we have the other inequality $\,[F:H]\geq p^n\,$
