When \(\displaystyle k\geq 11, k!\equiv 0 \ \mbox{(mod 11)}\)

Since we are adding up to 11!, can this be done without \(\displaystyle \sum_{i=1}^{10}k!\) individual?

\(\displaystyle 1!+2!+\dots +10!\equiv x \ \mbox{(mod 11)}\)

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