Is the finite sum of factorials constant modulo the summation limit?

Let $ K ( q ) = \sum _ { k = 1 } ^ { q - 1 } k ! $. Since $ q ! \equiv 0 \pmod q $, thus $ c $ is an integer such that for every positive integer $ q $, we have $ K ( q ) \equiv c \pmod q $. for instance, we have $ K ( q ! ) \equiv c \pmod { q ! } $. But by definition of $ K $, we know that $ K ( q ! ) \equiv K ( q ) \pmod { q ! } $, which leads to $ K ( q ) \equiv c \pmod { q ! } $. Hence there is a sequence of integers like $ ( k _ q ) _ { q \in \mathbb Z ^ + } $ such that $ c = k _ q \cdot q ! + K ( q ) $. Now for every positive integer $ q $: $$ 0 = c - c = k _ { q + 1 } \cdot ( q + 1 ) ! + K ( q + 1 ) - k _ q \cdot q ! - K ( q ) = \big( ( q + 1 ) k _ { q + 1 } - k _ q + 1 \big) q ! $$ $$ \therefore \quad k _ { q + 1 } = \frac { k _ q - 1 } { q + 1 } $$ Now using induction we show that for every natural number $ n $, we must have $ | k _ q | \ge q ^ n $. For the base case, we note that if $ k _ q = 0 $, then $ k _ { q +1 } $ can't be an integer, so $ | k _ q | \ge 1 = q ^ 0 $. For the induction step, we have: $$ \frac { | k _ q | + 1 } { q + 1 } \ge \frac { | k _ q - 1 | } { q + 1 } = | k _ { q + 1 } | \ge ( q + 1 ) ^ n $$ $$ \therefore \quad | k _ q | \ge ( q + 1 ) ^ { n + 1 } - 1 \ge q ^ { n + 1 } $$ But this leads to an obvious contradiction. So $ c $ doesn't exist.