Note that to get the bound $\textrm {rank}\left(t\right)\leq n^{k-1}$ for $t\in F^n\otimes ...\otimes F^n$, suppose that
$$t=\sum _{i=1}^r u_i \otimes v_i$$
with $u_1,...,u_r\in F^n\otimes ...\otimes F^n$ themselves rank-one with $k-1$ copies of $F^n$, and $v_1,...,v_r\in F^n$. Suppose that $r$ is the minimal number for such a decomposition. We will show that $u_1,...,u_r$ is linearly independent, and hence that $r\leq n^{k-1}$.
Suppose for contradiction that $u_1,...,u_r$ is linearly dependent
$$\sum _{i=1}^r a_iu_i=0$$
without loss of generality, assume that $a_r\neq 0$ and hence we may assume $a_r= -1$. We have
$$t=\sum _{i=1}^r u_i \otimes v_i$$
$$=\sum _{i=1}^{r-1} u_i \otimes v_i+u_r \otimes v_r$$
$$=\sum _{i=1}^{r-1} u_i \otimes v_i+\sum _{i=1}^{r-1} a_iu_i \otimes v_r$$
$$=\sum _{i=1}^{r-1} u_i \otimes \left(v_i+a_iv_r\right)$$
but this contradicts the minimality of $r$. Hence, $u_1,...,u_r$ is linearly independent, thus $\textrm {rank}\left(t\right)\leq n^{k-1}$.