I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$ S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1) $$ with the conditions that $S_q(0,k)=\delta_{0,k}$ and $S_q(n,0)=\delta_{n,0}$.
How is this recurrence arrived at? Even if this recurrence is taken as definition, there must be some motivation for it. Thank you.
On
If you can read German you find an answer in my lecture notes at University of Vienna, Elementare q-Identitäten, Chapter 3.
Your $S_q(n,k)$ is the same as my $q^{\binom{k}{2}} S[n,k]$
The following justification is based on material in Johann Cigler’s lecture notes, Chapters 1 and 3.
The ordinary Stirling numbers of the second kind are characterized by the identity $$(xD)^n=\sum_k\left\{\matrix{n\\k}\right\}x^kD^k\;,\tag{1}$$ where $D$ is the ordinary differentiation operator. Thus, one approach to defining a $q$-analogue $S_q(n,k)$ is to require that it satisfy the analogue of $(1)$ with $D$ replaced by $D_q$, defined by $$D_qf(x)=\frac{f(qx)-f(x)}{qx-x}$$ and satisfying $$D_qx^n=\frac{(qx)^n-x^n}{(q-1)x}=\frac{q^n-1}{q-1}x^{n-1}=(n)_qx^{n-1}$$ and $$D_qx=qxD_q\;.$$
In other words, one might reasonably define $S_q(n,k)$ to satisfy
$$(xD_q)^n=\sum_{k=0}^nS_q(n,k)x^kD_q^k\;.\tag{2}$$
Assume that $(2)$ holds for some $n$; then
$$\begin{align*} xD_q(xD_q)^n&=\sum_kS_q(n,k)xD_qx^kD_q^k\\ &=\sum_kS_q(n,k)x\Big(q^kx^kD_q+(k)_qx^{k-1}\Big)D_q^k\\ &=\sum_kS_q(n,k)q^kx^{k+1}D_q^{k+1}+\sum_k(k)_qS_q(n,k)x^kD_q^k\\ &=\sum_k\left(S_q(n,k-1)q^{k-1}+(k)_qS_q(n,k)\right)x^kD_q^k\;, \end{align*}$$
so if we want $(2)$ to hold for $n+1$, we must set
$$S_q(n+1,k)=S_q(n,k-1)q^{k-1}+(k)_qS_q(n,k)\;.\tag{3}$$
$(2)$ clearly requires that $S_q(n,0)=\delta_{n,0}$; it imposes no constraint on $S_q(0,k)$ for $k>0$, but setting it to $0$ is the natural thing to do and is compatible with $(3)$.