Show that $\lbrace 1, \alpha^k, \alpha^{2k}, \cdots\rbrace$ span $\ell^2$ for $0<|\alpha|<1$ and $k \geq 1$

To clarify, not every element of $\ell^2$ is the sum of finitely many $f_k$; take an element of $\ell^2$ that does not decay as $e^{-tn}$ for some $t$, for example. It is true, however, that the closed span of the $f_k$ (i.e., the closure of the vector space they generate) is $\ell^2$ itself.

Fix $x = (x_0, x_1, \dots)\in \ell^2$, and assume without loss of generality that each $x_i$ lies in $X = [-1, 1]$. Since $x_i \to 0$, there exists (e.g., by the Tietze extension theorem) some continuous $f:X \to X$ with $f(\alpha^n) = x_n$ for each $x$. Fix $\epsilon > 0$, and choose $N$ such that $$\sum_{n > N} |x_n|^2 < \epsilon.$$ By the Stone-Weierstrass theorem, there exists some polynomial $g(z) = \sum a_n z^n$ with $|g - f| < \epsilon$ on $X$. Then $\xi = \sum a_k f_k\in \ell^2$ has $$|\xi_n - x_n| = \left|\sum_k a_k \alpha^{nk} - x_n\right| = |g(\alpha^n) - x_n| < \epsilon$$ for all $n$. Now bound the $\ell^2$-norm of $\xi - x$, using the fact that the $(f_k)_n$ decay exponentially in $n$.

Recall that a set $S\subseteq \ell^2$ is dense if $({\bf c},s)=0$ for all $s\in S$ implies ${\bf c}=0$.

Fix $\alpha$ such that $0<|\alpha|<1$ and let $$S=((1,\alpha^k,\alpha^{2k},\dots):k=1,2\dots)\subset \ell^2.$$

Let ${\bf c}\in \ell^2$ be such that $({\bf c},s)=0$ for all $s\in S$ and define

$$f(x) = ({\bf c},(1,x,x^2,\dots)) = \sum_{j=0}^\infty c_j x^j.$$

Then $f$ is analytic on $(-1,1)$ (and by extension on the open unit ball in the complex plane). Since by assumption $f(\alpha^k)=0$ for all $k=1,2,...$, and $\alpha^k \to 0$ as $k\to\infty$, it follows from the uniqueness theorem for analytic functions that $f$ is identically zero. Therefore, $c_k = \frac{f^{(k)}(0)}{k!}=0$, proving that ${\bf c}=0$. Thus $S$ is dense.