Reciprocal of Square of 1 Less than Number Base

Theorem
Let $b \in \Z$ be an integer such that $b > 2$.

Let $n = \paren {b - 1}^2$.

The reciprocal of $n$, expressed in base $b$, recurs with period $b - 1$:


 * $\dfrac 1 n = \sqbrk {0 \cdotp \dot 012 \ldots c \dot d}_b = \sqbrk {0 \cdotp 012 \ldots cd012 \ldots}_b$

where:
 * $c = b - 3$
 * $d = b - 1$

Proof
By Basis Representation Theorem, the number $\sqbrk {12 \ldots cd}_b$ can be written as:

Thus the recurring basis expansion $\sqbrk {0 \cdotp \dot 012 \ldots c \dot d}_b$ can be written as:

Hence the result.