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Find a simplified form for $n!+(n−1)!+(n−2)!+(n−3)!+\dots+1!$ .

By simplified, I mean that there should not be "..." in the equation.

If there isn't one, prove it. Thanks!

Edit 1: Is it possible not to have $\sum$ in it?

Michael Hardy
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asdf
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    This may not be satisfactory but: http://mathworld.wolfram.com/FactorialSums.html Also another way to say simplify is "in closed form". – Kieran Cooney Oct 29 '13 at 23:16
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    $\sum_{a=1}^n a!$ has no "..." in it. – hmakholm left over Monica Oct 29 '13 at 23:17
  • Here have some forms http://www.wolframalpha.com/input/?i=sum%28a%21%2C+a%3D1...n%29 – Integral Oct 29 '13 at 23:34
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    You can use the gamma function, turn that expression into a single integral by linearity, and use the geometric sum to avoid having a $\sum$ in the integrand... that avoids the $\sum$ but, is that simplfied? not really... you'd get $$ \int_0^\infty e^{-t} \cdot \left(\frac{t^{n+1}-t}{t-1}\right) dt $$ – Pablo Rotondo Oct 29 '13 at 23:36
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    Even a single summand didn't have a simple form until the "$!$" notation was introduced! – Casteels Oct 29 '13 at 23:45
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    Although the Mathworld article pointed out by Kieran already mentions it, I'd like to explicitely reference http://oeis.org/A007489 here. – MvG Apr 03 '14 at 12:29

2 Answers2

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This number is just $$1111\cdots1111$$ with $n+1$ digits in factorial number system.

Berci
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∑(2-k)!

This is a simplified form of n!+(n−1)!+(n−2)!+(n−3)!+...+1!

Adrian Santarossa
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