We present a new explicit formula for the determinant that contains superexponentially fewer terms than the usual Leibniz formula. As an immediate corollary of our formula, we show that the tensor rank of the $n \times n$ determinant tensor is no larger than the $n$-th Bell number, which is much smaller than the previously best-known upper bounds when $n \geq 4$. Over fields of non-zero characteristic we obtain even tighter upper bounds, and we also slightly improve the known lower bounds. In particular, we show that the $4 \times 4$ determinant over ${\mathbb{F}}_2$ has tensor rank exactly equal to $12$. Our results also improve upon the best-known upper bound for the Waring rank of the determinant when $n \geq 17$, and lead to a new family of axis-aligned polytopes that tile ${\mathbb{R}}^n$.

We show that the product rank of the $3\,\times \,3$ determinant ${{\det }_{3}}$ is $5$ , and the product rank of the $3\,\times \,3$ permanent $\text{per}{{\text{m}}_{3}}$ is $4$ . As a corollary, we obtain that the tensor rank of ${{\det }_{3}}$ is $5$ and the tensor rank of $\text{per}{{\text{m}}_{3}}$ is $4$ . We show moreover that the border product rank of $\text{per}{{\text{m}}_{3}}$ is larger than $n$ for any $n\,\ge \,3$ .