I am not aware of any other argument. Conceptually it is actually quite simple. The main idea consists in changing the viewpoint and considering, instead of the simple random walk on a single tree, the random walk on the whole ensemble of Galton-Watson trees. In other words, one has to pass from a countable state space (the set of vertices of a single tree) to the continuous state space $\mathcal T$ which consists of (the isomorphism classes) of all pointed ($\equiv$ rooted) trees (this is a tree together with a distinguished vertex). There is a naturally defined simple random walk on $\mathcal T$, which consists in replacing the root of a tree $\tau\in\mathcal T$ with one of its neighbours with equal probabilities, and the key ingredient of the proof is the observation that the simple random walk on $\mathcal T$ has a stationary probability measure $\pi$ naturally related to the original branching process.

The original Galton-Watson measure is not stationary because the simple random walk is not stochastically homogeneous with respect to it: the root is an orphan and therefore has fewer neighbours than the offspring. However, this drawback is easy to repair; one has to add "by force" one more offspring to the first generation, and to let it multiply at the following steps in the same way as the other offspring. Let $\pi$ be the resulting measure on $\mathcal T$. The measure $\pi$ can also be described by saying that it is obtained by running the branching process not from a single progenitor, but from two progenitors joined with an edge.

The fact that the measure $\pi$ is stationary implies, by the subadditive ergodic theorem, existence of a rate of escape $\ell$ for the simple random walk on $\pi$-a.e. tree. Therefore, by the above description of the measure $\pi$, the rate of escape for the simple random walk on a.e. Galton-Watson tree also exists and equals $\ell$.

Let
$$
\ell_n = \mathbf E_\pi d(x_0,x_n)
$$
be the expectation, with respect to the stationary measure $\pi$, of the distance between the positions $x_0$ and $x_n$ of the simple random walk at times 0 and $n$, respectively. Then $\ell=\lim \ell_n/n$ by the definition of $\ell$. However, the limit of the differences $\ell_{n+1}-\ell_n$ also exists (and therefore must coincide with $\ell$).

For finding $\lim(\ell_{n+1}-\ell_n)$ one has to notice that $\ell_{n+1}$ coincides, by the stationarity of $\pi$, with the expectation of $d(x_{-1},x_n)$. Now, if $x_n\neq x_0$ and the degree of $x_0$ is $\delta$, then the expectation of $d(x_{-1},x_n)-d(x_0,x_n)$ is precisely $(\delta-2)/\delta$. Therefore, by transience of a.e. Galton-Watson tree (I will skip an explanation), one obtains that
$$
\ell = \mathbf E_\pi \frac{\delta-2}{\delta}.
$$
Since the distribution of the degree $\delta$ of the root with respect to the measure $\pi$ is $p'_k=p_{k-1}$, this is precisely the formula you are asking about.

Random discrete structures (Minneapolis, MN, 1993), 185–198, IMA Vol. Math. Appl., 76, Springer, New York, 1996. dx.doi.org/10.1007/978-1-4612-0719-1_12 pages.iu.edu/~rdlyons/ps/dimTalk.ps You might like to start by reading that. $\endgroup$ – Nate Eldredge Aug 20 '16 at 15:56