partial sequences of coin flips converge to uniform$[0,1]$ 
Let $\{B_i\}$ be iid Bernoulli$(\frac{1}{2})$ random variables and consider the number in $[0,1]$ whose binary expansion corresponds to the infinite sequence $\{B_i\}_{i=1}^{\infty}$, i.e. the number
  \begin{align*}
\sum_{n=1}^{\infty} 2^{-n} B_n.
\end{align*}
  Let $X_n$ be the $n^{th}$ partial sum, i.e.
  \begin{align*}
\sum_{k=1}^{n} 2^{-k}B_k.
\end{align*}
  Show that $X_n$ converges in distribution to a continuous Uniform$[0,1]$ random variable.

To show convergence in distribution, we can show that the CDF's of the $\{X_n\}$, which we can denote $\{F_n\}$, converge to the CDF of Uniform$[0,1]$, which is $F(x) = x$.
I tried to compute $F_n(x)$, which is
\begin{align*}
P(X_n \leq x) &= P(\sum_{k=1}^n 2^{-k}B_k \leq x) \\
&= P(\sum_{k=1}^n 2^{n-k} B_k \leq 2^n x) 
\end{align*}
I don't know how to simplify this expression.
Another approach I took was to compute the characteristic function of $X_n$ and show that it converges to the characteristic function of the uniform$[0,1]$ random variable. After some calculations, this amounted to showing that
\begin{align*}
2^{-n} \prod_{k=1}^{n} (1 + e^{2^{-k}it}) \to \frac{e^{it} - 1}{it}
\end{align*} 
for each $t \in \mathbb{R}$. Again I am stuck here. Can anyone find a slick way to prove this intuitive result?
 A: $$P\left(X_n < \frac{1}{2^k}\right) = \frac{2^{n-k}}{2^n}= \frac{1}{2^k}$$
The only way $X_n$ can be less than $\frac{1}{2^k}$ is when $B_{i}=0, \forall i \leq k$ and $\forall j > k, B_j$ can take either $0$ or $1$.
$$\sum_{i=k+1}^{n}\frac{1}{2^i} < \sum_{i=k+1}^{\color{red}{\infty}}\frac{1}{2^i} = \frac{1}{2^k}$$
The corresponding probability of $X_n < \frac{1}{2^k}$ will then be the ratio of number of possible binary strings of length $n-k$ to the number of possible binary strings of length $n$.
I am not sure how to proceed further with mathematical equations. But,
Another way to prove the result is:
$\forall x \in [0,1]$, there exists a bijective mapping between $x$ and binary strings (of possibly infinite length) where $0 \leftrightarrow 000\ldots$ and $1 \leftrightarrow 111\ldots$. For a fixed $x_0$, the binary representation be $b_1b_2\ldots b_n$. The fraction of the number of binary strings which evaluate to a number less than equal to $x_0$ and greater than equal to $0$, will equal to ratio of the length of the interval $[0,x_0]$ to the total length of the interval $[0,1]$ which will be $x_0$. This argument proves that $P(X_n\leq x) = x$.
