Show that if ${\frak h}\oplus x\Bbb F$ is solvabe and $e_0$ is a simultaneous eigenvector, then $\pi(x)^pe_0$ are all in the same eigenspace

Question

This is a step in proving that any finite-dimensional representation of a solvable Lie algebra $\mathfrak g$ over an algebraically closed field has a simultaneous eigenvector, as discussed in these pdf notes.

Let $\mathfrak g$ be solvable, and let $\mathfrak h$ be a (solvable) ideal in $\mathfrak g$ of codimension one, $\mathfrak g=\mathfrak h\oplus x\mathbb F$ for some $x\in\mathfrak g$. Let $e_0\in V$ be a simultaneous eigenvector for $\mathfrak h$: $$\pi(h)e_0 = \lambda(h)e_0,\qquad\forall h\in\mathfrak h.$$ I want to show that this implies that $\pi(h)\pi(x)^p e_0=\lambda(h)\pi(x)^p e_0$ for all $p\ge0$ and $h\in\mathfrak h$.

The argument given in the above notes, as far as I understand it, goes roughly as follows:

1. Define $e_p\equiv \pi(x)^p e_0$.
2. Observe that $\pi(h)e_1=\lambda(h)e_1 + \pi([h,x])e_0$, and thus more generally $$\pi(h) e_p = \lambda(h) e_p \bmod{(e_0,e_1,...,e_{p-1})}.$$
3. Define $W\equiv \mathrm{span}(\{e_0,e_1,...\})$, and observe that $$\operatorname{Tr}_W\pi(h) = \lambda(h) \dim(W), \\ \qquad \operatorname{Tr}_W \pi([h,x]) = 0 \Longleftrightarrow \lambda([h,x])=0.$$
4. ????
5. Thus $\pi(h)e_p=\lambda(h)e_p$ for all $h\in\mathfrak h$ and $p\ge0$.

They comment that this is shown by induction remembering the definition of $\lambda(h)$, but I'm not quite seeing this. I suppose I need to use the trace properties above into $\pi(h)e_1=\lambda(h)e_1 + \pi([h,x])e_0$ to realise that $\pi([h,x])=0$. It's probably really simple but I'm not seeing it.