If you do not want to use the row number explicitly from the position index $i$ then you can use the inefficient $$\frac{\Big\lfloor \dfrac{\sqrt{8i+1}-1}{2}\Big\rfloor!}{\left(i-\dfrac{\big\lfloor \frac{\sqrt{8i+1}-1}{2}\big\rfloor\big\lfloor \frac{\sqrt{8i+1}+1}{2}\big\rfloor}{2}\right)!\left(\Big\lfloor \dfrac{\sqrt{8i+1}-1}{2}\Big\rfloor+\dfrac{\big\lfloor \frac{\sqrt{8i+1}-1}{2}\big\rfloor\big\lfloor \frac{\sqrt{8i+1}+1}{2}\big\rfloor}{2}-i\right)!}$$
That is just a substituted version of a more efficient method of finding the row $r=\Big\lfloor \dfrac{\sqrt{8i+1}-1}{2}\Big\rfloor$ using the floor function and and the value of the index of the initial position in the row $s=\dfrac{r(r+1)}{2}$, so the value corresponding to index position $i$ is $\displaystyle {r\choose i-s}=\dfrac{r!}{(i-s)!(r+s-i)!}$
With respect to the comments following your query, with both $A$ and $k$ fixed, I don't know of any way of precisely computing the largest value of $n$ such that $\binom{n}{k} \leq A.$ However, I think a lower bound and an upper bound for the precise value of $n$ can be supplied, purely from the definition of $\binom{n}{k}.$
Let $B = A \times k!$.
Let $g(n) \equiv (n) \times (n-1) \times \cdots \times (n - [k-1]).$
Then you want the largest $n$ so that $g(n) \leq B.$
$g(n) < n^k,$ so if $n$ is chosen so that $n^k < B$,
then this guarantees that $g(n) < B.$
$n^k < B \iff k \log n < \log B \iff \log n < \frac{\log B}{k}.$
Therefore, a lower bound for $n$ is given by
choosing the largest $n$ such that $\log n < \frac{\log B}{k}.$
Similar analysis can be used to provide an upper bound for $n$.
Let $m = (n - [k-1]).$
Clearly, $g(n) > m^k.$
Further, if $\log m > \frac{\log B}{k}$
Then $\log m^k = k \times \log m > \log B ~\Rightarrow $
$m^k > B.$
Therefore, an upper bound for $n$ is given by
choosing the smallest $n$ such that $\log (n - [k-1]) > \frac{\log B}{k}.$
Note, that the above approach merely provides a starting point for (perhaps) attaining tighter bounds. Tighter bounds can be sought by trying to more accurately determine the geometric mean of $g(n).$
That is, you are looking for some value $r$ where $[n - (k-1)] < r < n$
and $r^k = g(n).$
It is known that the geometric mean of (for example) the integers $i$
such that $ [n - (k-1)] \leq i \leq n$
is less than the arithmetic mean of the integers in this range
(i.e. the average $V_n ~= \frac{n + [n - (k-1)]}{2}).$
This means that the first part of the above analysis, which used the idea
that $n^k > g(n)$
may instead use the idea that $(V_n)^k > g(n)$.
Repeating the analysis re the search for a lower bound for $n$
this means that a tighter (i.e. larger) lower bound can be computed
by choosing the largest $n$ such that $\log (V_n) < \frac{\log B}{k}.$
However, we have now reached the boundary of where my knowledge can offer any support.
Best Answer
This is OEIS A001142 which begins $$1, 1, 2, 9, 96, 2500, 162000, 26471025, 11014635520, 11759522374656, 32406091200000000, 231627686043080250000, 4311500661703860387840000, 209706417310526095716965894400, 26729809777664965932590782608648192$$ An approximate formula is given $a(n) \approx A^2 * \exp(n^2/2 + n - 1/12) / (n^{(n/2 + 1/3)} * (2*\pi)^{((n+1)/2))}$, where $A = A074962 = 1.2824271291\ldots$ is the Glaisher-Kinkelin constant.
The growth rate is dominated by the term $$\frac{\exp(\frac {n^2}2)}{n^{\frac n2}}=\exp \left( \frac{n^2}2-\frac n2\log n \right)$$