Why is the tensor product of complex planes $\mathbb C$ again just $\mathbb C$?

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As the title says already, why does the following equality hold

\begin{equation} \bigotimes^N \mathbb C = \mathbb C \end{equation}

My question comes from an example regarding the bosonic Fock space $\mathcal F_+$ over $\mathbb C$ and the harmonic oscillator where we had

\begin{equation} \mathcal F_+(\mathbb C) = \bigoplus\limits_{N = 0}^\infty \bigotimes\limits_{\text{sym}}^N \mathbb C = \bigoplus\limits_{N = 0}^\infty \mathbb C \end{equation} My Professor said that tensorizing one dimensional spaces gives the space itself. I kind of get what is meant but not how this applies here. Thanks for any answere in advance.

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It is a basic fact (which you can find in every text on tensor products, for example here) that the tensor product of $K$-vector spaces $\otimes_K$ satisfies $$K \otimes_K V \cong V$$ for every $K$-vector space $V$. The isomorphism is defined by using the universal property of the tensor product via the $K$-bilinear map $K \times V \to V$, $(a,v) \mapsto av$. Thus, it maps $a \otimes v \mapsto av$. The inverse map is $v \mapsto 1 \otimes v$.

In particular, we have $K \otimes_K K \otimes_K \cdots \otimes_K K \cong K$ for any finite number of copies of $K$. Notice that it is not precise to write equality here.