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).

(First note Gadi's comment below the question.)
In my previous answer, I derived the inequality
$$
{(2d-1)n-1 \choose n-1} \leq \frac{1}{{2d - 1}}\bigg[\frac{{(2d - 1)^{2d - 1} }}{{(2d - 2)^{2d - 2} }}\bigg]^n ,
$$
which is already more than needed because of the factor $1/(2d-1)$ (recall that in the case $d=2$ the right-hand side is equal to $(1/3)6.75^n$). However, this inequality can be further improved to 
$$
{(2d-1)n-1 \choose n-1} \leq \bigg[\frac{{(2d - 1)^{2d - 1} }}{{(2d - 2)^{2d - 2} }}\bigg]^{n-1} ,
$$
which is optimal since equality holds when $n=1$.
The last inequality can be proved by induction as follows (for $d \geq 2$ a fixed integer). First, we have equality when $n=1$. Next note that
$$
{(2d-1)(n+1)-1 \choose n} = \frac{{((2d - 1)n - 1)!}}{{(n - 1)!((2d - 2)n)!}}\frac{{(2d - 1)n}}{n}\frac{{\prod\nolimits_{k = 1}^{2d - 2} {((2d - 1)n + k)} }}{{\prod\nolimits_{k = 1}^{2d - 2} {((2d - 2)n + k)} }}.
$$
Since, for any $1 \leq k \leq 2d-2$, 
$$
\frac{{(2d - 1)n + k}}{{(2d - 2)n + k}} \le \frac{{2d - 1}}{{2d - 2}},
$$
it thus follows that
$$
{(2d-1)(n+1)-1 \choose n} \leq {(2d-1)n-1 \choose n-1} \frac{{2d - 1}}{1} \bigg(\frac{{2d - 1}}{{2d - 2}}\bigg)^{2d - 2} .
$$
Now, under the induction hypothesis that
$$
{(2d-1)n-1 \choose n-1} \leq \bigg[\frac{{(2d - 1)^{2d - 1} }}{{(2d - 2)^{2d - 2} }}\bigg]^{n-1} ,
$$
we get
$$
{(2d-1)(n+1)-1 \choose n} \leq \bigg[\frac{{(2d - 1)^{2d - 1} }}{{(2d - 2)^{2d - 2} }}\bigg]^{n},
$$
completing the proof.
