$\DeclareMathOperator\lead{leader} \DeclareMathOperator\prob{prob}$Answering a follow-up question by Per Alexandersson. Here is the $q$-version obtained by a suitable modification of the probabilistic proof of the OP identity.
We consider the linear space $X:=\mathbb{F}_q^n$ over a finite field $\mathbb{F}_q$. For $x=(x_1,\ldots,x_n)\in X\setminus {0}$ denote $\lead(x)=\max(i:x_i\ne 0)$, for a subspace $L\subset X$, $m:=\dim L>0$, denote $\lead(L)=\max_{x\in L} \lead(x)$. It follows from Gauss elimination that $L$ contains a basis $f_1,\ldots,f_m$, such that $\lead(f_1)<\lead(f_2)<\ldots <\lead(f_m)=\lead(L)$. Thus $L$ contains exactly $q^m-q^{m-1}$ elements $x$ for which $\lead(x)=\lead(L)$, and $q^{m-1}$ 1-dimensional subspaces $R$ for which $\lead(R)=\lead(L)$.
Choose a random subspace $L$ of $X:=\mathbb{F}_q^n$ with probability of $k$-dimensional subspace proportional to $q^{k\choose 2} y^k$ ($k=1,2,\ldots,n$). The sum of these weights is $(1+y)(1+qy)\ldots (1+q^{n-1}y)-1:=\theta_n$ (that's $q$-binomial theorem).
Then choose a random 1-dimensional subspace $R\subset L$ uniformly. Consider the following probability: $\kappa:=\prob(\lead(R)=\lead(L))$. On one hand,
$$
\kappa=\sum_{k=1}^n \prob(\dim L=k)\prob(\lead R=\lead L|\dim L=k)\\
=\theta_n^{-1}\sum_{k=1}^n q^{k\choose 2}y^k{n\choose k}_q\cdot \frac{q^{k-1}}{[k]_q},
$$
where $[k]_q=1+q+\ldots+q^{k-1}$ is the number of 1-dimensional subspaces of a $k$-dimensional space over $\mathbb{F}_q$.
On the other hand, denoting by $X_k$ the $k$-dimensional subspace of $x\in X$ for which $x_{k+1}=\ldots=x_n=0$, we get
$$
\kappa=\sum_{k=1}^n \prob(L\subset X_k\& \lead(R)=k)=\sum_{k=1}^n \prob(L\subset X_k)\cdot \prob(\lead(R)=k|L\subset X_k)\\=
\theta_n^{-1}\sum_{k=1}^n \theta_k\cdot \frac{q^{k-1}}{[k]_q}.
$$
Thus the identity (we multiply both expressions for $\kappa$ by $q\cdot \theta_n$)
$$
\sum_{k=1}^n q^{k+1\choose 2}y^k{n\choose k}_q\cdot \frac{1}{[k]_q}=\sum_{k=1}^n \frac{q^{k}((1+y)(1+qy)\ldots (1+q^{k-1}y)-1)}{[k]_q},
$$
for $q=y=1$ we get the initial identity.
Best Answer
Tonight I read here [the answer by esg to another your question] that $\frac1{2\pi}\int_{-\pi}^\pi e^{-ik t}(1+e^{it})^ndt=\binom{n}{k}$, which is, well, obvious at least when both $n$ and $k$ are positive integers: just expand the binomial $(1+e^{it})^n$ and integrate. Denoting $\alpha=k/n$ we may rewrite this as $\frac1{2\pi}\int_{-\pi}^\pi (f(t))^n dt=\binom{n}{\alpha n}$, where the function $f(t)=(1+e^{it})e^{-i\alpha t}$ is complex-valued. For making it real-valued, we change the path between the points $-\pi$ and $\pi$. The value of the integral does not change (since $f^n$ is analytic between two paths, for integer $n$ it is simply entire function.) On the second path $f$ takes real values. Namely, for $t\in (-\pi,\pi)$ we define $s(t)=\ln \frac{\sin (1-\alpha)t}{\sin \alpha t}$. It is straightforward (some elementary high school trigonometry) that $$f(t+is(t))=\frac{\sin t}{\sin^{\alpha} \alpha t\cdot \sin^{1-\alpha}(1-\alpha)t},$$ so we replace the path from $(-\pi,\pi)$ to $\{t+s(t)i:t\in (-\pi,\pi)\}$ (limit values of $s(t)$ at the endpoints are equal to 0) and take only the real part of the integral (this allows to replace $d(t+s(t)i)$ to $dt$ in the differential). We get $$ \frac1{2\pi}\int_{-\pi}^\pi \frac{\sin^n t}{\sin^{\alpha n} \alpha t\cdot \sin^{(1-\alpha)n}(1-\alpha)t}dt=\binom{n}{\alpha n} $$ as desired.