Consider $(-1)^{\dfrac{(p-1)(p-2)}{2}}$. For $p = 2n$ we have
\begin{align}
(-1)^{\dfrac{(p-1)(p-2)}{2}} &= (-1)^{\dfrac{(2n-1)(2n-2)}{2}} \\
&= (-1)^{(2n-1)(n-1)} \\
&= (-1)^{2n^2 - n + 2n + 1} \\
&= i^{2(2n^2 + n + 1)} \\
&= i^{4n^2 + 2n + 2} \\
&= - i^{4n^2 + 2n} \\
&= - i^{2n} \\
&= - (i^2)^{n} \\
&= - (-1)^{n} \\
&= (-1)^{n+1} \\
&= (-1)(-1)^{n} \\
&= (-1)(i^2)^{n} \\
&= - i^p
\end{align}
and for $p = 2n + 1$ we have
\begin{align}
(-1)^{\dfrac{(p-1)(p-2)}{2}} &= (-1)^{\dfrac{(2n+1-1)(2n+1-2)}{2}} \\

Let $g$ be defined on $\mathbb R$ by $g(1):=0$, and $g(x):=2$ if $x \neq 1$, and let $f(x):=x + 1$ for all $x\in \mathbb R$. Show that $\lim_{x\to 0}g \circ f \neq g( f(0))$. Why doesn’t this contradict Theorem 5.2.6?
5.2.6 Theorem Let $A, B \subset \mathbb R$ and let $f : A \to \mathbb R$ and $g : B \to \mathbb R$ be functions such that
$f (A) \subset B$. If $f$ is continuous at a point $c\in A$ and $g$ is continuous at $b=f(c)\in B$, then the
composition $g \circ f : A \to \mathbb R$ is continuous at $c$.