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I am trying to prove the following statement:
If $G$ is a torsion group and nilpotent of class $c$, then $\exp(G)$ divides $\exp(G^{\operatorname{ab}})^c$ (where $\exp$ is the exponent, i.e. the lcm of the order of the group elements, and $G^{\operatorname{ab}}$ is the abelianization of $G$).
I thought about using induction on $c$. The base case is easy: for $c=1$, $G = G^{\operatorname{ab}}$, so the statement follows. I have no idea how to do the induction step though.
Consider the lower central series of $G$, $G_1=G$, $G_2=[G,G]$, $G_{n+1} = [G_n,G]$. Since $G$ is of class $c$, $G_{c+1}=\{e\}$. Let $N=\mathrm{exp}(G^{\rm ab})$.
We have that $G/G_2\cong G^{\rm ab}$ has exponent $N$. Thus, for every $x\in G$, $x^N\in G_2$.
Let us look next at $G/G_3$. If $x\in G$, then $x^N\in G_2$. Now, $G_2$ is generated by elements of the form $[r,s]$ with $r\in G_1$, $s\in G$; and in $G_2/G_3$ we have $$[r,s]^N \equiv [r^N,s]\pmod{G_3}$$ and $[r^N,s]\in [G_2,G]=G_3$ so $[r,s]^N\in G_3$. And since $G_2/G_3$ is abelian, generated by elements of exponent $N$, it is itself of exponent $N$.
Thus, if $x\in G$, then $x^N\in G_2$, and $x^{N^2}=(x^N)^N\in G_3$.
Now look at $G_3/G_4$. This is an abelian group; since $G_3$ is generated by elements of the form $[g,s]$ with $g\in G_2$, then $G_3/G_4$ is generated by their images modulo $G_4$. But we have that $$[g,s]^N \equiv [g^N,s]\pmod{G_4}$$ and $g^N\in G_3$, since $g\in G_2$. Therefore $[g,s]^N\in [G_3,G]=G_4$. Thus, for every $x\in G$, $x^{N^3}\in G_4$.
Continuing this way, we obtain that for every $x\in G$, $x^{N^c}\in G_{c+1}=\{e\}$, as desired.
You can probably frame this as a “descending induction”, by starting with elements in $G_c$ and proving they are of exponent $N$; then showing that if the elements of $G_{c-k}$ are of exponent $N^{k+1}$, with $k\lt c-1$, then the elements of $G_{c-(k+1)}$ are of exponent $N^{k+2}$; and thus concluding that the elements of $G= G_1 = G_{c-(c-1)}$ is of exponent $N^{(c-1)+1}$.
