Showing that Hadamard is its own inverse
Rewrite your final expression as
$$
\sum_{z \in \{0,1\}^n} \left| z \right\rangle \cdot \frac{1}{2^n} \sum_{y \in \{0,1\}^n} (-1)^{y \cdot (x+z)}.
$$
So to complete the proof, we need to show that
$$
\sum_{y \in \{0,1\}^n} (-1)^{y \cdot (x+z)} = \begin{cases} 2^n &amp; \text{if } x=z, \\ 0 &amp; \text{otherwise}. \end{cases}
$$
If $x=z$ then $x+z = 0$ (computing modulo 2!), and so $(-1)^{y \cdot (x+z)} = 1$ for all $y$, hence the sum equals $2^n$. If $x \neq z$ then pick some $i$ such that $x_i \neq z_i$. We can partition $y$ into the $i$th coordinate $y_i$ and all the rest $y_{-i}$. Now
$$
\sum_{y \in \{0,1\}^n} (-1)^{y \cdot (x+z)} =
\sum_{y_{-i} \in \{0,1\}^{n-1}} (-1)^{y_{-i} \cdot (x+z)_{-i}} \sum_{y_i \in \{0,1\}} (-1)^{y_i \cdot (x+z)_i} = 0,
$$
since $x_i+z_i = 1$ (modulo 2) and $\sum_{y_i \in \{0,1\}} (-1)^{y_i} = 1-1=0$.