Bounding ${(2d-1)n-1\choose n-1}$ Claim: ${3n-1\choose n-1}\le 6.25^n$.


*

*Why? 

*Can the proof be extended to
obtain a bound on ${(2d-1)n-1\choose
    n-1}$, with the bound being $f(d)^n$
for some function $f$?
(These numbers describe the number of some $d$-dimensional combinatorial objects; claim 1 is the case $d=2$, and is not my claim).

In the meanwhile, you may consider the following.
Suppose that $X$ is a binomial$((2d-1)n-1,p)$ random variable ($0 < p < 1$). Then,
$$
{\rm P}(X = n - 1) = {(2d-1)n-1 \choose n-1} p^{n-1}(1-p)^{n(2d-2)} \leq 1.
$$
Putting $z=1/p > 1$, we thus have
$$
{(2d-1)n-1 \choose n-1} \leq z^{n-1} \bigg(\frac{z}{{z - 1}}\bigg)^{n(2d - 2)} = \frac{1}{z}\bigg[\frac{{z^{2d - 1} }}{{(z - 1)^{2d - 2} }}\bigg]^n .
$$
So, we want to minimize 
$$
\psi(z) = \frac{{z^{2d - 1} }}{{(z - 1)^{2d - 2} }},\;\; z > 1.
$$
In the case where $d=2$, we get
$$
{3n-1 \choose n-1} \leq \frac{1}{z}\bigg[\frac{{z^3 }}{{(z - 1)^2 }}\bigg]^n. 
$$
Here, $\psi(z) = \frac{{z^3 }}{{(z - 1)^2 }}$ attains its minimum at $z=3$, where $\psi(z)=27/4$.
Hence
$$
{3n-1 \choose n-1} \leq \frac{1}{3}\bigg(\frac{{27}}{4}\bigg)^n = \frac{1}{3}6.75^n .
$$
EDIT: For general $d$, the function $\psi(z)$ attains its minimum at $z=2d-1$ (indeed, $\psi(z) \to \infty$ as $z \downarrow 1$ or $z \to \infty$, and, as an elementary calculation shows, $\psi'(z)=0$ for $z=2d-1$). Hence,
$$
{(2d-1)n-1 \choose n-1} \leq \frac{1}{{2d - 1}}\bigg[\frac{{(2d - 1)^{2d - 1} }}{{(2d - 2)^{2d - 2} }}\bigg]^n .
$$
