Tensor product of Hadamard operators The Hadamard Operator on one qubit is:
\begin{align*}
H = \tfrac{1}{\sqrt{2}}\left[\,\left(\color{darkgreen}{|0\rangle + |1\rangle}\right)\color{darkblue}{\langle 0|}+\left(\color{darkgreen}{|0\rangle - |1\rangle}\right)\color{darkblue}{\langle 1|}\,\right]
\end{align*}
Show that:
\begin{align*}
H^{\otimes n} = \frac{1}{\sqrt{2^n}}\sum_{x,y}(-1)^{x \cdot y}\,\left|x\rangle \langle y\right|
\end{align*}
I can evaluate things like $H \otimes H$ in practice, but I don't know how to get a general formula for $H^{\otimes n}$. Are there any tricks I could use?
 A: The Keyword here is mathematical induction: Suppose that the formula holds for some $n$ and show that therefore it holds for $n+1$. If you additionally show that it holds for $n=2$, you have shown the general formula for arbitrary $n$. 
A: This can be shown directly. The definition you posted can be seen as the single-qubit version of your target,
$$
H=\frac{1}{\sqrt{2}}\sum_{x_1,y_1}(-1)^{x_1 y_1}| x_1 \rangle\langle y_1 |.
$$
The $n$-qubit case is then
$$
\begin{align}
H^{\otimes n} &=\frac{1}{\sqrt{2^n}}\left(\sum_{x_1,y_1}(-1)^{x_1 y_1}| x_1 \rangle\langle y_1 |\right) \otimes\cdots\otimes \left(\sum_{x_n,y_n}(-1)^{x_n y_n}| x_n \rangle\langle y_n |\right) 
\\& =\frac{1}{\sqrt{2^n}}\sum_{x_1,\ldots,x_n \\ y_1,\ldots,y_n}(-1)^{x_1 y_1}\cdots(-1)^{x_n y_n}| x_1 \rangle \otimes\cdots\otimes | x_n \rangle\langle y_1 | \otimes\cdots\otimes \langle y_n |
\end{align}
$$
by expanding the sum. The only difference between this and your goal, 
$$
H^{\times n}=\frac{1}{\sqrt{2^n}}\sum_{x,y}(-1)^{x \cdot y}| x \rangle\langle y |,
$$
is notation.
