Im reading this paper in the Direct counting argument section(I'm only interested in this method of proof) and it is said that once we have proven $d=3$, the $d>3$ case follows as a generalisation left as an exercise. I don't see how it does, could someone explain?
My idea was that it suffices to show that the first three coordinates are not "recurrent" in some sense, i.e. we can look in one plane only. Since it is not recurrent that we will return to the origin of this plane, we're done. But I'm not convinced by this because when $d$ increases, we divide by $k^n$ with $k$ larger (See proof), so the sum might actually end up converging.
Let $X_n$ denote your random walk on $\mathbb Z^d$ ($d > 3$). Restrict it by looking at only the first three coordinates. Delete any entries where those first three coordinates are stationary (indicating that the movement for that $n$ would have occurred in one of the discarded dimensions). You now have a genuine simple random walk on $\mathbb Z^3$. Because that random walk will only visit $(0,0,0)$ finitely many times, the original random walk $X_n$ will only visit $(0, 0, 0, \dots, 0)$ finitely many times as well.
EDIT: The above is, to me, the conceptually cleanest way to see the link between the transience of $\mathbb Z^3$ and the higher-dimensional lattices. However, upon inspecting the paper a bit more closely, I think the paper was hinting for a more combinatorial approach. The basic idea would be to consider a higher-dimensional expression this term that appears on page 6: $$u_{2n} = \left( \frac 1 6 \right)^{2n} \sum_{\star} \frac{(2n)!}{L!L!U!U!(n-U-L)!(n-U-L)!}$$ then to use Stirling's approximation to show that this term is asymptotically less than $\frac{M}{n^{d/2}}$. In four dimensions, this would look something like (adjustments in blue): $$u_{2n} = \left( \frac 1 {\color{blue}{8}} \right)^{2n} \sum_{\color{blue}{\star}} \frac{(2n)!}{L!L!U!U!\color{blue}{Z! Z!}(n-U-L-\color{blue}{Z})!(n - U-L-\color{blue}{Z!})}$$ to account for the new cardinal direction.