Given $7^{45}\equiv32 \bmod 101$, find $32^{-1}$ using square and multiply.
What I did was since $(7^{45})^{-1}$ is equal to $(32)^{-1}$, and $-45\bmod 101 = 56$ then $32^{-1} = 7^{56}$. I then used square and multiply but I get $16$ for $7^{56}$ but I know that the inverse of $32$ modulo $101$ is $60$.
Am I missing a property of square and multiply or of modular arithmetic for this not to work?
People have pointed you to the right direction. I'll just do it here for my own sake. Everything below is $\pmod{101}$.
$$ 32^{-1}\equiv 7^{55}\equiv 108^{55}\equiv (2^2 \cdot 3^3)^{55}\equiv 2^{110} \cdot 3^{165} \equiv 2^{10} \cdot 3^{65} \equiv 2^{10} \cdot 3 \cdot (81)^{16} \\ \equiv 2^{10} \cdot 3 \cdot (-20)^{16} \equiv 2^{10} \cdot 3 \cdot 2^{16} \cdot 10^{16} \equiv 2^{26} \cdot 3 \cdot 100^8 \equiv \frac{(2^9)^3 \cdot 3}{2}\\ \equiv \frac{512^3}{2}\cdot 3 \equiv \frac{7^3}{2} \cdot 3 \equiv \frac{343}{2}3 \equiv \frac{40}{2}\cdot 3 \equiv 60. $$