Further Congruency within the Nottingham Group 
Write $A$ for the group of continuous automorphisms of $\mathbb{F}_{p}[[t]]$ and for each $n \geq 1$ let $J_{n}$ be the kernel of the homomorphism from $A$ to the automorphism group of $\mathbb{F}_{p}[[t]]/(t^{n+1})$, where $(t^{n+1})$ is the ideal generated by $t^{n+1}$.
For each $n \geq 1$ let $e_{n} \in J_{n}$ be defined by $e_{n}(t) = t + t^{n+1}$. Assume for simplicity that $p \neq 2$. Prove that $[e_{r}, e_{s}] \equiv e_{r+s}^{r-s} \mod J_{r+s+1}$ for all $r, s \geq 1$. Deduce that $J_{1}$ is generated by $\{e_{1}, e_{2}\}$.

This is a continuation on my previous question, Congruency within the Nottingham Group, which shows that $J_{1}$ can be generated by $\{e_{i} \mid i \geq n\}$. I've managed to prove the first part of this exercise but I'm struggling to make the following deduction. My effort is as follows:
$[e_{n}, e_{1}] \equiv e_{n+1}^{n-1} \mod J_{n+2}$. Then we can recursively express powers of $e_{n}$ for $n \geq 3$ purely in terms of powers of $e_{1}, e_{2}$ modulo $J_{n+1}$.
I'm not really convinced by my own argument and I'm way out of my depth to be honest, any help would be appreciated.

You are missing some technical details. First of all, the group $J_1$ is a profinite group, so the proper meaning of $\{ e_1, e_2 \}$ generating it is that it generates a dense subgroup. In the category of profinite groups, this means that $\{ e_1, e_2 \}$ generates all of the quotients $J_1/J_n$ for $n \geq 2$. These are all finite $p$-groups, and your recursive argument works well in this setting. 
