The rank is one provided that $u\otimes v \neq 0$. In particular, $u, v \neq 0$.
Consider the matrix
$$A = \left[\begin{array}{cccc}
r_1s_1 & r_1s_2 & \dots &r_1s_n\\
r_2s_1 & r_2s_2 & \dots & r_2s_n\\
\vdots & \vdots & \ddots&\vdots\\
r_ms_1 & r_ms_2 & \dots & r_ms_n
\end{array}\right].$$
Note that
\begin{align*}
\operatorname{rank}(A) &= \dim\operatorname{Col}A\\
&= \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1s_1\\ \vdots\\ r_ms_1\end{array}\right], \dots, \left[\begin{array}{c} r_1s_n\\ \vdots\\ r_ms_n\end{array}\right]\right\}\\
&= \dim\operatorname{span}\left\{s_1\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right], \dots, s_n\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\
&= \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\
&= 1
\end{align*}
as $u = \sum r_iu_i$ where $u_i$ form a basis and $u \neq 0$, so $r_i \neq 0$ for some $i$.
Another way of seeing it uses the notion of outer product (see here and here). Let $v^T = [r_1\ \dots\ r_m]$ and $w^T = [s_1\ \dots\ s_n]$, then the outer product of $v$ and $w$ is
$$vw^T = \left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right][s_1\ \dots\ s_n] = \left[\begin{array}{cccc}
r_1s_1 & r_1s_2 & \dots &r_1s_n\\
r_2s_1 & r_2s_2 & \dots & r_2s_n\\
\vdots & \vdots & \ddots&\vdots\\
r_ms_1 & r_ms_2 & \dots & r_ms_n
\end{array}\right] = A.$$
As $A$ can be written as the sum of one outer product and no fewer (because $A$ is not the zero matrix), $A$ has rank one.