Does $\text{Span}(\{1, e^x, e^{2x}, . . . e^{nx}, . . .\})=C(\mathbb{R})$?

Hint: What kind of bounded functions are spanned by those functions?

The Laplace transform of a function in $\text{Span}(1,e^x,e^{2x},e^{3x},\ldots)$ may only have simple poles at the non-negative integers, but $$ \mathcal{L}\left(2^{-x}\right) = \frac{1}{s+\log 2}$$ and $\log(2)$ is both positive and irrational, hence $2^{-x}\not\in \text{Span}(1,e^x,e^{2x},e^{3x},\ldots)$. A similar argument applies to $\sin(x)$: we have $\mathcal{L}(\sin x)=\frac{1}{1+s^2}$ with simple poles at $\pm i\not\in\mathbb{R}$, from which $\sin(x)\not\in\text{Span}(1,e^x,e^{2x},e^{3x},\ldots)$. In $\text{Span}(1,e^x,e^{2x},e^{3x},\ldots)$ we don't even have the identity function, since the Laplace transform of $x$ has a double pole at the origin.

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However, if we replace $\mathbb{R}$ with a sub-interval $I=[a,b]$ we get something interesting. If we consider a continuous (or differentiable) function $f(x)$ over $I$, we may define $g(x)=f(\log x)$ over the interval $J=[e^a,e^b]$. $g(x)$ is continuous over $J$, hence by the Weierstrass approximation theorem there is a sequence of polynomials $p_n(x)$ uniformly convergent to $g(x)$ on $J$, and the sequence $p_n(e^x)\in\text{Span}(1,e^x,e^{2x},\ldots)$ provides a uniform approximation of $f(x)$ over $I$.
For instance, over the interval $\left[-\frac{1}{2},\frac{1}{2}\right]$ we have $$ x\approx -\frac{11}{6}+3e^x-\frac{3}{2}e^{2x}+\frac{1}{3}e^{3x} $$

This proves that $\text{Span}(1,e^x,e^{2x},\ldots)$ is dense in $C([a,b])$.

The span of the functions $e^{nx}$, $n\in{\mathbb N}_{\geq0}$, consists of the finite linear combinations of these functions, and not a single function more than that. In other words, it consists of the functions $x\mapsto p\bigl(e^x\bigr)$, with $p$ an arbitrary polynomial of one variable.