Show that $7^6 \equiv 2^7 \pmod{223}$. Given $7 \cdot2^5 \equiv 1\pmod{223}$. Show that $7^6 \equiv 2^7 \pmod{223}$.
I know there must be some clever way to show this congruence. I can't seem to figure it out. I've considered that $224=7\cdot2^5$ and tried multiplying both sides of the first congruence by various terms, but with little success. Any solutions or hints are greatly appreciated.

A not clever method?
$$
\begin{aligned} 7^6 &\equiv (7^3)^2 \\
                    &\equiv 343^2 \\
                    &\equiv 120^2 \\
                    &\equiv 240 \times 60 \\
                    &\equiv 17 \times 60 \\
                    &\equiv 60 + 4 * 4 * 60 \\
                    &\equiv 60 + 4 * 17 \\
                    &\equiv 128 \ ({\rm mod}\ 223)
\end{aligned}
$$
