Fix $n$ and consider $P_k(x)=(1+\alpha)(1+2\alpha)\cdots(1+(k-1)\alpha)\cdot{x\choose n}$, where $\alpha$ is the operator sending ${x\choose k}$ to ${x\choose k-1}$. That is, $\alpha$ sends $f(x)$ to $f(x+1)-f(x)$. We'll show by induction that each $P_k(x)$ has all real roots and that the distance between consecutive roots is at least $1$. The claim is trivial for $k=1$.

Assume the claim holds for $P_k(x)$. Then $P_{k+1}(x)=(1+k\alpha)P_k(x)$ is the linear combination $kP_k(x+1)-(k-1)P_k(x)$. Let the roots of $P_k(x)$ be $a_1<\cdots<a_n$. The roots of $P_k(x+1)$ are $a_1-1<\cdots<a_n-1$, which interlace with those of $P_k(x)$ by the induction assumption. In the notation of Fisk's "Polynomials, roots, and interlacing", $P_k(x+1)\underline\ll P_k(x)$. Let the roots of $P_{k+1}(x)$ be $b_1\le\cdots\le b_n$. It follows from (the proof of) Prop 1.35 that all the roots are real and that $b_1\le a_1-1$ and $a_{i-1}\le b_i\le a_i-1$ for all $i>1$. Therefore $b_i-b_{i-1}\ge a_{i-1}-(a_{i-1}-1)=1$, which completes the induction.