We want to find $M={180}^{59}mod391$.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{180}^2}& =32400=82\times 391+338\\ \multicolumn{1}{c}{}& \equiv 338mod391.\end{array}}$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{180}^3}& =5832000=14915\times 391+235\\ \multicolumn{1}{c}{}& \equiv 235mod391.\end{array}}$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{180}^{59}}& =(180){}^2({180}^3){}^{19}\\ \multicolumn{1}{c}{}& \equiv 338\times {235}^{19}\\ \multicolumn{1}{c}{}& \equiv ({235}^2){}^9\times 235\times 338\\ \multicolumn{1}{c}{}& \equiv (55225){}^9\times 79430\\ \multicolumn{1}{c}{}& \equiv (141\times 391+94){}^9\times (203\times 391+57)\\ \multicolumn{1}{c}{}& \equiv {94}^9\times 57\\ \multicolumn{1}{c}{}& \equiv {830584}^3\times 57\\ \multicolumn{1}{c}{}& \equiv (2124\times 391+100){}^3\times 57\\ \multicolumn{1}{c}{}& \equiv 57000000\\ \multicolumn{1}{c}{}& \equiv 20mod391\end{array}}$