The digit in unit's place of number $\frac{{10}^{2013}-1}{{10}^{33}-1}$ is ________

$\begin{array}{c}\frac{{10}^{2013}-1}{{10}^{33}-1}=\frac{{\left({10}^{33}\right)}^{61}-1}{{10}^{33}-1}\\ =\left({\left({10}^{33}\right)}^{60}+{\left({10}^{33}\right)}^{59}+{\left({10}^{33}\right)}^{58}+\dots +\left({10}^{33}\right)+1\right)\end{array}$

Thus, digit in unit place is 1