Let H be the Hadamard operator, prove $H^{\otimes n} \left| 0 \right>^{\otimes n} = \sum_{i=0}^{2^n -1} \left| i \right>$ Let H be the Hadamard operator.
 $$ H = (\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right| -\left| 1 \right> \left< 1 \right| )$$
prove that. $$H^{\otimes n} \left| 0 \right>^{\otimes n} = \sum_{i=0}^{2^n -1} \left| i \right>$$ 
Now, it is evident that this works for $n=1$ and $n=2$, because we know that.
$$ \left| 0 \right>^{\otimes 2} = \left| 0 \right> \otimes \left| 0 \right> = \left| 00 \right> $$
Then.
$$ H^{\otimes2}\left| 00 \right> = (\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right| -\left| 1 \right> \left< 1 \right| )(\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right| -\left| 1 \right> \left< 1 \right| )\left| 00 \right>$$
$$H^{\otimes2}\left| 00 \right> = \frac{1}{2}(\left< 00 \right|\left| 00 \right>\left| 00 \right> + \left< 00 \right|\left| 00 \right>\left| 01 \right> + \left< 00 \right|\left| 00 \right>\left| 10 \right> + \left< 00 \right|\left| 00 \right>\left| 11 \right>)$$
$$H^{\otimes2}\left| 00 \right> = \frac{1}{2}(\left| 00 \right> + \left| 01 \right> + \left| 10 \right> + \left| 11 \right>) = \frac{1}{2}\sum^{3}_{i=0}\left| i \right> = \frac{1}{\sqrt{2^2}}\sum^{2^2 -1}_{i=0}\left| i \right> $$
but I do not know how to proof that it works for $n = k + 1$.
Any clue is very welcome, thank you in advance for your time and advices. 
For more information about  bra-ket  notation, or the  Hadamard  operator you can consult those links. 
 A: Perhaps it suffices to show that when represented as a matrix in the standard basis, the first row of $H_n := H^{\otimes n}$ is just a row containing all $1$'s, or when properly normalized, containing all $\frac{+1}{2^{n/2}}$.  This follows directly as a consequence of $H_n$ being the tensor product $H_n = H_{1} \otimes H_{n-1}$.  See https://en.wikipedia.org/wiki/Hadamard_transform and follow the steps to arrive at the formula for the entries of $H_n$ as
\begin{equation}
(H_n)_{i,j} = \frac{(-1)^{i\cdot j}}{2^{n/2}}
\end{equation}
where $i = 0$ for the upper row, and $i \cdot j$ is the bitwise dot product. 
Relating back to the question at hand, the notation $H^{\otimes n} | 0 \rangle^{\otimes n}$ means ``take the coefficients of the first row of $H_n$, and weight the states $|j\rangle$ by their respective entry, for $j=0,1,\dots,2^n-1$."  Since each entry of this row is $\frac{1}{2^{n/2}}$, this will be precisely the RHS that you want to show.  Therefore, the crux of the proof is really showing that $H_n = H_1 \otimes H_{n-1}$, from which the formula for $(H_n)_{i,j}$ follows.
