For each \( n \geq 7\), \(n!\) is divisible by \(14\) (as it is divisible by both \(2\) and \(7\)), so we only have to consider the first six terms:
\[ 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 = 62 \cdot 14 + 5, \]
so the answer is \(\boxed{5}\).

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TopNewestFor each \( n \geq 7\), \(n!\) is divisible by \(14\) (as it is divisible by both \(2\) and \(7\)), so we only have to consider the first six terms: \[ 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 = 62 \cdot 14 + 5, \] so the answer is \(\boxed{5}\).

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