Surprisingly, although this is stated everywhere in literature (usually citing surveys), there seems to be no recent source that explicitly proved this statement (there are, however, sources that proved this for a general case where there is damping factor, but the proof cannot be adapted to non-damping scenario).
To define the original problem, suppose that a graph is undirected and connected. Define $d(i)$ as the degree centrality (aka degree) of node $i$, $\mathbf{A}$ the adjacency matrix. Let $\pmb{w}$ denote the PR vector, and consider PageRank where there is no damping factor, i.e.
$$w(i)=\sum_{j}{A_{ij}\frac{w(j)}{d(j)}}$$
Now we want to prove that $w(i)\propto d(i)$, or $\pmb{w}=\alpha \mathbf{D}\cdot\mathbf{1}$, where $\mathbf{D}$ is the degree diagonal matrix.
I know Perron-Fronebius theorem can tackle this, but I am thinking if there could be a more straightforward proof, possibly solving directly the PageRank equations.
Edit 1:
It looks like Perron-Fronebius cannot directly get the work done because we don't have a strictly positive matrix. Perhaps should consider proof by contradiction with Chapman-Kolmogorov equations if proving uniqueness.
Best Answer
I believe we can simply input $w = \alpha \cdot D \cdot \mathbf{1}$ into the equation and check that it works. Note that your equation $$w(i) = \sum_{j} A_{i, j} \frac{w(j)}{d(j)}$$ may be rewritten in matrix terms as: $$\pmb{w}^T = \pmb{w}^T\mathbf{D}^{-1}A$$ Transposing both sides: $$\pmb{w} = A\mathbf{D}^{-1}\pmb{w},$$ as both $A$ and $\mathbf{D}$ are symmetric. Let us insert $\pmb{w} = \alpha \mathbf{D}\cdot \mathbf{1}$ on the right side. $$\pmb{w} = A\mathbf{D}^{-1}\alpha \mathbf{D}\cdot \mathbf{1}$$ $$\pmb{w} = \alpha A \cdot \mathbf{1}$$ Note that $A\cdot\mathbf{1}$ is a vector of degrees, i.e. $(A \cdot\mathbf{1})_i = \sum_j A_{i, j} = d(i)$. So, we got that $$\pmb{w} = \alpha \textbf{D}\cdot1,$$ which confirms our assumption.