oh, that is clear for me. Correct my implications: if $5^{22}\equiv1\pmod{23}$ then $(5^{22})^{71119}\equiv1^{71119}\pmod{23}$ and $1^{71119}=1$, so $(5^{22})^{71119}\equiv1\pmod{23}$, then $(5^{22})^{71119}\cdot5^6\equiv1\cdot5^6\pmod{23}$, so $(5^{22})^{71119}\cdot5^6\equiv5^6\pmod{23}$
Right?