Fun in 2016

\[ \large \displaystyle\sum _{ { n }_{ 1 },{ n }_{ 2 },\ldots,{ n }_{ 2016 }=1 }^{ \infty }{ \dfrac { 1 }{ ({ n }_{ 1 }+{ n }_{ 2 }+\cdots+{ n }_{ 2016 }-1)! } } =\dfrac { e }{ a! } \]

Find \(a\).

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