Is $F^F$ a subspace of $F^{\infty}$ where $F$ is a field? 
Is $F^F$ a subspace of $F^{\infty}$ where $F$ is a field?

I am trying to prove that $F^\infty$ is an infinite-dimensional vector space by showing that its subspace is not finite-dimensional. I know that $\mathcal{P}(F)$ is infinite-dimensional and I also know that $\mathcal{P}(F)$ is a subspace of $F^F$. Hence it would be to sufficient to show that $F^{\infty}$ is infinite-dimensional if $F^F$ is a subspace of it. I am kind of new to these kinds of proofs, and I am a little fuzzy with the intuition behind $F^F$ and $F^\infty$, so I am stuck with my proof at this point.
Note that $\mathcal{P}(F)$ is the set of all polynomials.
 A: First, there is more than one "infinity", so let's fix an infinite set $I$ and talk about $F^I$ rather than $F^\infty$.
The easiest way to show that $F^I$ is infinite dimensional is to exhibit an infinite linearly independent family of vectors. Consider e.g. the family of vectors $(e_i)_{i\in I}$ defined as follows: the $i^\textrm{th}$ component of $e_i$ is $1$, all other components are $0$. In other words, if you write $e_i= (e_i^j)_{j \in I}$ then $e_i^j = \delta_{i, j}$ where $\delta_{i, j}$ is the Krönecker symbol.
All you have to do is to show that the family $(e_i)_{i\in I}$ is linearly independent, ie that whenever one considers pairwise distinct $i_1,\dots, i_n \in I$  and $\lambda_1, \dots, \lambda_n \in F$ satisfying $\sum_{k=1}^n \lambda_k e_{i_k} = 0$, then all $\lambda_k$s are equal to $0$. 
A: In general, $F^F$ is not a subspace of $F^\infty=F^\mathbb N$.
A simple example is $F=\mathbb R$. This is a set of cardinality $2^{\mathfrak c}$, where $\mathfrak c$ is the cardinality of $\mathbb R$, while $\mathbb R^\mathbb N$ has only cardinality $\mathfrak c$. Clearly a subspace cannot be larger than the full space.
Indeed, $F^F$ is (isomorphic to) a subspace of $F^\mathbb N$ iff $F$ is countable (that is, either finite or equinumerous to the natural numbers).
