Is $x_n(t) = t^n$ for $n\ge 1$ dense in $L^1[0,1]$? My question concerns the following problem

Let $x_n(t) = t^n$ for $n\ge 1$. Is $\mathrm{span}(x_n; n\ge 1)$ dense in $L^1([0,1])$?

By Weierstrass' approximation theorem, it suffices to check whether the constant function $1$ is contained in the $L^1$-closure of the linear span of the $x_n$. 
I think this is not the case, but I haven't found a proof so far.
Some simple observations: I know that if $1$ is contained in the $L^1$-closure, then we can find a sequence of polynomials $p_n\in \mathrm{span}(x_n; n\ge 1)$ such that $p_n\to 1$ in $L^1$, almost everywhere and almost uniformly. Furthermore the sequence $q_n(x) = \int_0^x p_n(t) \, dt$ will converge uniformly to $x$ on $[0,1]$.
I don't see how this helps, though.
 A: This is a write up of the solution based on t.b.'s nice idea in the comments:

Claim: Let $x_n(t) = t^n$. Then $\mathrm{span}(x_n; n\ge 1)$ is dense as a subspace of $L^1([0,1])$.

Proof: By monotone convergence, the sequence $x^{1/m}$ converges to $1$ in $L^1([0,1])$. It therefore suffices to check that $x^{1/m} \in \overline{\mathrm{span}(x_n; n\ge 1)}$ for all $m\in \mathbb N$. 
The latter follows from the fact that the sequence of Bernstein polynomials
$$B_k(x) =  \sum_{i=0}^k \binom ki \left(\frac{i}{k}\right)^{1/m}x^i (1-x)^{k-i}$$
converges to $x^{1/m}$ uniformly and is contained in $\mathrm{span}(x_n; n\ge 1)$ (the constant term where $i=0$ vanishes). $\square$
A: You can remove any finite collection of  terms from the series of monomials $(t^n)$ ($n=0,1,2,3,\dots$) and the resulting linear span is still dense in $C([0,1]$ (and so in the corresponding $L^1$-space).  This is a consequence of a  deep result---the theorem of Muentz---one of the  central results of approximation theory.  You can easily find information and the precise formulation (which allows one to remove suitable infinite sets of indices) on Wikipedia.
A: Approximating $1$ by polynomials without constant term is essentially the same as approximating $1/x$ by arbitrary polynomials.
Of course there is a twist in that $1/x$ is not continuous on $[0,1]$, but take $P(x)$ to approximate $\min\{1/x,M\}$ for some large $M$ to within $\epsilon$ (in sup norm).
Then $|xP(x)-1| < \epsilon$ for all $x > 1/M$, and for $x < 1/M$ we have $0 < P(x) < 2M$ (assuming $\epsilon$ is reasonable) so $|xP(x)-1| \le 1$.  The $L^1$ difference is bounded by $\epsilon + 1/M$, which we can send to $0$.
