On a theorem of Burnside The following is a well known theorem of Burnside, which I am reading from the original paper.
If $G$ is a finite $p$-group and $\{S_1,S_2,\cdots, S_n\}$ is a set of elements of $G$ such that their images in $G/[G,G]$ is a (minimal) generating set for $G/[G,G]$ then $\{ S_1,S_2,\cdots, S_n\}$ is a minimal generating set for $G$ .
I know a proof of this theorem which uses some characterization of the Frattini subgroup of a $p$-group. But, I will be happy if one can explain the original proof of Burnside (see link). I came at this proof by a comment of Philip Hall, who himself couldn't clarify the proof of Burnside.
 A: This follows Jack's comment. I write in modern group theory symbols: First, consider the case that $G'$ is abelian.(I think this is the key point). 
Consider $\bar{G}=G/G'$. Let $\bar{s}_1, \cdots, \bar{s}_n$ be the minimal generating set of $\bar{G}$, where $s_i \in G$. Let $S=\langle s_1, \cdots s_n \rangle$. Then $G=SG'$. Let $N=S\cap G'$. Since $G'$ is abelian in this case, we get $N \unlhd G$. Now we can consider $\tilde{G}=G/N$. Now $\tilde{G}=\tilde{S}\ltimes \tilde{G}'$. Let $\tilde{K}$ be a subgroup of index $p$ in $\tilde{G}'$. We can even choose $\tilde{K}$ to be normal in $\tilde{G}$ for $G$ is a $p$-group. Let $\tilde{L}=\tilde{S} \tilde{L}$, then $\tilde{G}/\tilde{L}$ is order $p$, and $\tilde{G}' \le \tilde{L}$, contrary the choice of $\tilde{L}$. 
Now the general case: As above we get $s_i$ and $S$. Suppose $G'' \ne 1$. Let $\bar{G}=G/G''$. Non $\bar{s}_i$ and $\bar{S}$ work by above. So we get $\bar{G}=\bar{S}$, i.e. $G=SG''$. In this way, we can get $G=SG^{(3)},\cdots$. But $G$ is $p$-group, and $G^{(m)}=1$ for some $m$. So we complete this proof.
