Hexadecimal magic fraction

Claim: in number base $b$ , $\frac 1{(b-1)^2} = 0.\overline{01234\cdots(b-4)(b-3)(b-1)}01234\dots.$

Substituting $b = 16$ gives the denominator $(16-1)^2 = 225 = \boxed{E1}_{16}.$

Proof: Let $a$ be the repeating part: $a = \overline{1234\cdots(b-4)(b-3)(b-1)} = 1 + \sum_{n=1}^{b-2} n\cdot b^{b-2-n}.$ Mulitply this by $b-1$ : $a(b-1) = (b-1) + \sum_{n=1}^{b-2} n(b-1)\cdot b^{b-2-n} \\ = (b-1) + \sum_{n=1}^{b-2} n \cdot b^{b-1-n} - \sum_{n=1}^{b-2} n \cdot b^{b-2-n} \\ = (b-1) + \sum_{n=1}^{b-2} n \cdot b^{b-1-n} - \sum_{n=2}^{b-1} (n-1) \cdot b^{b-1-n} \\ = (b-1) + \sum_{n=1}^{b-1} \left(n-(n-1)\right) \cdot b^{b-1-n} - (b-1) \\ = \sum_{n=1}^{b-1} b^{b-1-n} = \sum_{n=0}^{b-2} b^n;$ multiply again by $b-1$ : $a(b-1)^2 = \sum_{n=0}^{b-2} (b-1) \cdot b^n \\ = \sum_{n=1}^{b-1} b^n - \sum_{n=0}^{b-2} b^n \\ = b^{b-1} - 1.$ The repeated fraction is $x = 0.00\dots00100\dots00100\dots\ \cdot\ a = \left(\sum_{p=1}^\infty b^{-p\cdot(b-1)}\right)\cdot a,$ so that $x(b-1)^2 = 0.00\dots00100\dots00100\dots\ \cdot\ a (b-1)^2 \\ = 0.00\dots00100\dots00100\dots\ \cdot\ (b^{b-1} - 1) \\ = \sum_{p=1}^\infty b^{-p\cdot(b-1) + b-1} - \sum_{p=1}^\infty b^{-p\cdot(b-1)} \\ = \sum_{p=0}^\infty b^{-p\cdot(b-1)} - \sum_{p=1}^\infty b^{-p\cdot(b-1)} \\ = b^0 = 1.$ From $xb^2 = 1$ it follows that $1/b^2 = x$ .

In answer to the bonus question, it's $\frac{1}{(b-1)^2}$

Applying this to the main problem, n = 15² = 225. So in hexadecimal n = E1