For finite dimensional $V$ and $W$, we know that \begin{equation*} \dim{L(V,W)} = \dim{V}\cdot\dim{W} \end{equation*}
Does this theorem hold for infinite-dimensional vector spaces $V$ and $W$ too?
For finite dimensional $V$ and $W$, we know that \begin{equation*} \dim{L(V,W)} = \dim{V}\cdot\dim{W} \end{equation*}
Does this theorem hold for infinite-dimensional vector spaces $V$ and $W$ too?
The answer I linked in my comment shows in particular that if we write $|X|$ for the cardinality of a set $X$, then for a vector space $V$ over $K$: $$ |V| = \max(|K|,\dim(V)).$$
Now if $(e_i)_{i\in I}$ is a basis of $V$, then $L(V,W)$ is in bijection with the functions $I\to W$ (this is what it means to be a basis), so $$ |L(V,W)| = |W|^{\dim(V)}.$$
Since that cardinal is clearly strictly bigger than $|K|$ when $\dim(V)$ is infinite, we have $|L(V,W)| = \dim(L(V,W))$ so $$\dim(L(V,W)) = |W|^{\dim(V)}.$$
So now you have two cases:
Note that the behaviour of cardinal exponentiation can be a little tricky: see the problem of the continuum hypothesis. If you are comfortable with assuming the Generalized Continuum Hypothesis (I personally am), then https://en.wikipedia.org/wiki/Continuum_hypothesis#The_generalized_continuum_hypothesis gives you a complete list of formulas to compute exponentiations.