$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$
See Definition 3.1.2 for regular singular points.
Again we can assume that $z_0=0$. Let us multiply our \begin{equation} u^{(n)}(z)+ p_{n-1} (z)u^{(n-1)}(z)+\ldots+p_1 (z)u'(z)+p_0(z)u(z)=0 \label{eq-3.3.1} \end{equation} by $z^n$: \begin{equation} Lu:=z^n u^{(n)}(z)+q_{n-1} (z) z^{n-1} u^{(n-1)}(z)+\ldots+ q_1 (z) zu'(z)+q_0(z)u(z)=0 \label{eq-3.3.2} \end{equation} and according to Definition 3.1.2 \begin{equation} q_k(z) := z^{n-k}p_k(z)\qquad k=0,1,\ldots, n-1 \label{eq-3.3.3} \end{equation} are analytic functions at $0$. Let \begin{equation} \bar{q}_k := q_k(0)\qquad k=0,1,\ldots, n-1 \label{eq-3.3.4} \end{equation} and \begin{equation} \bar{L}u:=z^n u^{(n)}(z)+\bar{q}_{n-1} z^{n-1} u^{(n-1)}(z)+\ldots+ \bar{q}_1 zu'(z)+\bar{q}_0(z)u(z) \label{eq-3.3.5} \end{equation} and \begin{equation} L':=L-\bar{L}. \label{eq-3.3.6} \end{equation} Observe that \begin{equation} \bar{L}z^\alpha:=I(\alpha)z^\alpha,\qquad I(\alpha)=\sum_{0\le k \le n} \alpha (\alpha-1)\cdots (\alpha -k+1)\bar{q}_k. \label{eq-3.3.7} \end{equation} Therefore if $L=\bar{L}$ (the case of the Euler equation) $u=z^\alpha$ satisfies $Lu=0$ if and only if $I(\alpha)=0$.
Definition 1. We call $I(\alpha)$ incidical polynomial and its roots incidical exponents.
On the other hand, $L'z^\alpha$ contains only $z^{\alpha+1},z^{\alpha+2},\ldots$. Let $\alpha$ be an incidical exponent. Then $Lz^\alpha$ contains only $z^{\alpha+1},z^{\alpha+2},\ldots$: \begin{equation*} Lz^\alpha = \sum _{l\ge 1} c_l(\alpha) z^{\alpha+l}. \end{equation*} To compensate $z^{\alpha+1}$ we add to $z^\alpha$ correction $u_1z^{\alpha+1}$; then \begin{equation*} Lz^\alpha = I(\alpha+1) u_1 z^{\alpha+1} + \sum _{l\ge 1} c_l(\alpha) z^{\alpha+l} + \sum _{l\ge 1} c_l(\alpha+l+1) u_1 z^{\alpha+l+1} \end{equation*} and the coefficient at $z^{\alpha+1}$ is $I(\alpha+1)u_1+c_1(\alpha)$. Choosing \begin{equation*} u_1= - \frac{1}{I(\alpha)}c_1(\alpha) \end{equation*} we can get rid off $z^{\alpha+1}$. In a similar manner we can get rid of $z^{\alpha+l}$ for all $l=1,2,\ldots$ provided \begin{equation} I(\alpha+k)\ne 0 \qquad\forall l=1,2,\ldots \label{eq-3.3.8} \end{equation}
It is a condition of Subsection 3.1.3. So we proved the first statement here
Theorem 1. Let $I(\alpha=0)$ and (\ref{eq-3.3.8}) be fulfilled. Then
Proof of Statement 2. It could be done in the same manner as of Theorem 3.2.1. We skip it.
Assume now that $\alpha$ is an incidical exponent of multiplicity $r\ge 2$ which means that \begin{equation} I(\alpha)=I'(\alpha)=\ldots =I^{(r-1)}(\alpha)=0, \qquad I^{(r)}(\alpha)\ne 0. \label{eq-3.3.10} \end{equation} Then not only $u=z^\alpha$ satisfies $\bar{L}u=0$, but also $u=z^\alpha (\log z)^j$ for all $j=1,\ldots,r-1$ (but not for $j=r$).
But what about $L' z^{\alpha} (\log z)^j$? It contains all powers of logarithm $\le j$: \begin{equation*} L' z^\alpha (\log z)^j = \sum _{l\ge 1}\sum_{k\le j}c_{lk}(\alpha) z^{\alpha+l} (\log z)^k. \end{equation*} Also \begin{equation*} \bar{L} z^{\alpha+l}(\log z)^k = I(\alpha+l) z^{\alpha+l}(\log z)^k + \sum_{m\le k-1} d_{km} z^{\alpha+l} (\log z)^m. \end{equation*} Therefore in the same manner as above we can compensate term with $z^{\alpha+l}(\log z)^j$, then term with $z^{\alpha+l}(\log z)^{j-1}$ and so on up to term with $z^{\alpha+l}$ and then move to $l+1$ instead of $l$.
So we proved the first statement here
Theorem 2. Let (\ref{eq-3.3.10}) and (\ref{eq-3.3.8}) be fulfilled. Then for all $j=0,1,\ldots, r-1$
Proof of Statement 2. It could be done in the same manner as of Theorem 3.2.1. We skip it.
Now we get rid off assumption (\ref{eq-3.3.8}). At some moment we need to compensate term with $z^{\alpha+l}(\log z)^m$ (but we allow new terms with increased $l' > l$ instead of $l$ and $m'$ instead of $m$ and also terms with $m' < m$.
To do this we used $u_{lm}z^{\alpha+l}(\log z)^m$ but it worked only as $I(\alpha+l)\ne 0$. Now we assume that \begin{equation} I(\alpha+l)=I'(\alpha+l)=\ldots =I^{(r'-1)}(\alpha+l)=0, \qquad I^{(r')}(\alpha+l)\ne 0 \label{eq-3.3.13} \end{equation} with $r'\ge 1$.
One can prove that \begin{multline} \bar{L} z^{\alpha+l}(\log z)^{m'}= \frac{m'!}{r'!(m'-r')!} I^{(r')}(\alpha+l)z^{\alpha+l}(\log z)^{m'-r'}+\\ \sum_{k< m'-r'} d_{m'k}(\alpha+l) z^{\alpha+l}(\log z)^{k}\qquad \label{eq-3.3.14} \end{multline} and we can do the compensation at the expense of raising $m$ to $m'=m+r'$.
So we proved the first statement here
Theorem 3. Let (\ref{eq-3.3.10}) be fulfilled (with $r\ge 1$). Then for all $j=0,1,\ldots, r-1$
Proof of Statement 2. It could be done in the same manner as of Theorem 3.2.1. We skip it.
$\Leftarrow$ $\Uparrow$ $\Rightarrow$
if $\beta$ is not an incidical exponent its multiplicity is $0$. ↩