$0^{th}$ tensor power, $V^{\otimes 0} = \Bbb F$, definition, or mathematical construction? 95% sure I will be told it's just a definition, move on etc:

Is there mathematical reason why the $0^{th}$ tensor power is defined as:
$$V^{\otimes 0} = \Bbb F$$

Where $\Bbb F$ is the field that vectorspace $V$ lies over.

We define the $n^{th}$ tensor power of $V$ as:
$$ V^{\otimes n}=V\otimes V\otimes \cdots \otimes V$$
with basis:
$$\{v^{i_1}\otimes v^{i_2} \otimes \cdots \otimes v^{i_n}: 1\leq i_k \leq n\}$$

For some reason I feel like $V^{\otimes 0}$ should equal the trivial vectorspace.

So is it just a definition? If so, what is the motivation for this choice. Thanks.
 A: You want it to be true that $V^{\otimes(n+m)} \cong V^{\otimes n} \otimes V^{\otimes m}$, which means that $V^{\otimes 0}$ has to be the unit for the tensor product, which is the underlying field. This is a straightforward categorification of $a^0 = 1$. 
Alternatively, you can think of $V^{\otimes 0}$ as having basis the "empty tensor product" of zero vectors in $V$. It helps to think of this as the unit in the tensor algebra $T(V) = 1 \oplus V \oplus V^{\otimes 2} \oplus \dots $. 
A: That $\bigotimes^{0} V =\mathbb{F} $ can be shown using the universal property that defines an $n^\text{th}$ tensor product.
Universal Property.
Suppose $V$ is a vector space over $\mathbb{F}$. The $n^\text{th}$ tensor power $(\bigotimes^n V, \otimes_n )$ is up to isomorphism uniquely determined by the following universal property, for any vector space $W$ over  $\mathbb{F}$ and any $n$-linear mapping $\pmb{\varphi}: V^n \to W$, there
exists a unique linear mapping $\tilde{\pmb{\varphi}}:\bigotimes^n V \to W $ such that $\pmb{\varphi} = \tilde{\pmb{\varphi}} \circ \otimes_n$.
Preliminaries.
Suppose $V, W$ are vector spaces over $\mathbb{F}$. Then $V^0=\mathbb{F}^0 = \{0\}$
and every function from $\{0\}$ to $W$ (resp. $\mathbb{F}$) is a $0$-linear mapping (the defining conditions are emptily fulfilled) therefore
$$ \mathbf{Mlt}^0_\mathbb{F}(V,W) = \{ \hat{\pmb{w}}:\{0\}\to W, 0\mapsto \pmb{w} \ |\   \pmb{w}\in W\}\cong W,$$
$$ \mathbf{Mlt}^0_\mathbb{F}(V,\mathbb{F}) = \{ \hat{a}:\{0\}\to \mathbb{F}, 0\mapsto a\ |\  a\in \mathbb{F}\} \cong  \mathbb{F}.$$
On the other hand, any linear map from $\mathbb{F}$ to $W$ is given by $\pmb{w}\in W$ i.e.
$$ \mathbf{Hom}_ \mathbb{F}( \mathbb{F},W) = \{ \tilde{\pmb{w}}:  \mathbb{F} \to W, a \mapsto a\pmb{w}\ |\  \pmb{w}\in W\} \cong W.$$
Proposition.
Suppose $V$ is a vector space over $\mathbb{F}$. Then $(\mathbb{F}, \hat{1})$ has the universal property of the $0^\text{th}$ tensor power of $V$, i.e. $\bigotimes^0 V = \mathbb{F}$.
Proof. For every $\hat{\pmb{w}} \in \mathbf{Mlt}^0_\mathbb{F}(V,W)$, we have $\hat{\pmb{w}} = \tilde{\pmb{w}}\circ \hat{1}$ i.e. $$ (\tilde{\pmb{w}}\circ\hat{1})(0)  = \tilde{\pmb{w}}(\hat{1}(0))= \tilde{\pmb{w}}(1) = 1\pmb{w} =\pmb{w} =\hat{\pmb{w}}(0).$$
The same reasoning shows that the $0^\text{th}$ exterior power of $V$ is as well $\bigwedge^0 V=\mathbb{F}$. This is because $ \mathbf{Alt}^0_\mathbb{F}(V,W) = \mathbf{Mlt}^0_\mathbb{F}(V,W)$ and  $ \mathbf{Alt}^0_\mathbb{F}(V,\mathbb{F}) = \mathbf{Mlt}^0_\mathbb{F}(V,\mathbb{F})$ and $(\mathbb{F}, \hat{1})$ has as well the universal property of the $0^\text{th}$ exterior power of $V$.
