Prove that in the polynomial space $\mathbb R[X]$, the series $\{f_0, f_1, ..., f_n,...\}$ in which $\deg f_n = n, \forall n \in \mathbb N, f_0 \neq 0$ represents a basis.
I tried to give $f_k$ a general form and then write a polynomial with it, but I don't see how it can help.
On
Consider the finite dimensional subspaces $\mathcal{P}_n$ of polynomials in $\mathcal{P}:=\mathbf{R}[x]$ with degree at most $n$, with $n\ge 0$ integer. It is known that the dimension of each $\mathcal{P}_n$ is $n+1$ and that, by construction, $\bigcup_n \mathcal{P}_n=\mathcal{P}$. Now, fix a set $$ \mathscr{P}:=\left\{p_n: n \in \mathbf{N}\right\} \subseteq \mathcal{P} $$ such that $p_n \in \mathcal{P}_n \setminus \mathcal{P}_{n-1}$ for all $n\ge 1$ and $p_0 \neq 0$. By induction, it is easy to see that each finite set $ \left\{p_0,\ldots,p_n\right\} $ is a basis of $\mathcal{P}_n$. As a consequence $\mathscr{P}$ is a generator of $\mathcal{P}$. Finally, the elements of $\mathscr{P}$ are linearly independent. Hence $\mathscr{P}$ is a basis of $\mathcal{P}$.
You have to show that this set is spanning and any finite combination is linearly independent:
Suppose $\{f_{n_1} \dots f_{n_k} \}$ is a subset of your set. Then we have the following:
$$\sum_{i=1}^k \alpha_if_{n_i}=0$$ implies $\alpha_i=0 \forall i \in \{1, 2, \dots k\}$ since the $f$'s are polynomials and $0$ has degree $0$ so if you start from the highest degree (which exists since the set is finite) you can find that all $\alpha$'s are equal to $0$
Hence the set is linearly independent.
To show that it is spanning, you have to notice that every polynomial can be represented as a sum of elements of that set (possibly multiplied by a constant). But that is again true - start from the highest degree of your polynomial - call it $P(x)$ and suppose deg$(P(x))=n$. Substract a multiple of $f_n$ so that the degree of $P(x)-f_n$ becomes less than $n$ and proceed by induction on the degree.
Since those $2$ conditions are satisfied, this is indeed a basis