Let ${n\brace k}$ be the Strirling number of the second kind, such that ${n+1\brace k} =k{n\brace k}+{n\brace k-1}$ with ${0\brace 0}=1$. Let $j,p,n$ integers such that $1 \le j \le p \le n$.
I start with the following inequalities
$$ \tag{1} {p \brace j}{n \brace p} \ge {n-j \choose p-j}{n \brace j} $$
$$ \tag{2}{p \brack j}{n \brack p} \le {n-j \choose p-j}{n \brack j} $$
which are not difficult to prove by induction on $n$.
Then, replacing $n$ by $n+j$ and $p$ by $p+j$ and $j \ge 1$ unbounded, we have
$$ \tag{3} \frac{{p+j \brace j}{n+j \brace p+j }}{ n+j \brace j} \ge {n \choose p} \ge \frac{{p+j \brack j}{n+j \brack p+j }}{ n+j \brack j} .$$
Then I observed that these bounds for the binomial coefficient are increasingly tight when $j$ increases and it can be shown that both converge to ${ n \choose p}$ as $j$ tends to $\infty$. So I wanted to prove that
$$\cdot \cdot \frac{{p+j-1 \brace j-1}{n+j-1 \brace p+j-1 }}{ n+j-1 \brace j-1} \ge \frac{{p+j \brace j}{n+j \brace p+j }}{ n+j \brace j} \cdot \cdot \ge {n \choose p} \ge \cdot \cdot\frac{{p+j \brack j}{n+j \brack p+j }}{ n+j \brack j} \ge \frac{{p+j-1 \brack j-1}{n+j-1 \brack p+j-1 }}{ n+j-1 \brack j-1} \cdot \cdot$$
Then coming back to $n,p$ instead of $n+j,p+j$, we need to prove, for $1 \le j \le p \le n$ $$ \tag{4} {p-1 \brace j-1 }{n-1 \brace p-1 }{n \brace j}\ge{p \brace j }{n \brace p }{n-1 \brace j-1 } $$ $$\tag{5} {p-1 \brack j-1 }{n-1 \brack p-1 }{n \brack j}\le{p \brack j }{n \brack p }{n-1 \brack j-1 } .$$
Would a proof by induction on $n$ work for (4) and (5) ? That does not seem so easy. Or a combinatorial proof?
Remark: these inequalities would be quite strong statements. For instance in (4), after incrementation of $n,p,j$ by $1$, when we let $j =p-1$, we obtain $$ {n \brace p}^2 \ge {n \brace p+1}{n \brace p-1}\Big(\frac{p+1}{p}\Big)^2\frac{p}{p-1}+ \frac{2}{p(p-1)}{n \brace p-1}{n \brace p}$$ which is stronger than the known log-concavity of the Stirling numbers ${n \brace p}^2 \ge {n \brace p+1}{n \brace p-1}$.