Can n Divide !n ?

For any integer n the "left factorial function" is defined as 
!n = 0!+1!+2!+...+(n-1)!.   The conjecture that n does not divide !n 
for any n>2 was listed as B44 in R. Guy's "Unsolved Problems in Number
Theory" back in 1981.  In the 2nd edition of Guy's book, he says, "In 
a 91-03-21 letter, Reg. Bond offers an as yet unpublished proof of 
the conjecture."  I'm not aware that this claimed proof has ever 
been published.  Perhaps it's considered too trivial to be worth 
publishing(?)

In any case, while we await Reg.Bond's proof, here's a geometrical 
approach to the problem. Consider a p-dimensional Euclidean space 
with orthogonal coordinates, and take the two vectors

          X = [ 0!, 1!, 2!,..., (p-1)! ]

          Y = [ (p-1)!,..., 2!, 1!, 0! ]

The dot product (i.e., the scalar product) of X and Y is

           X.Y = (0!)(p-1)! + (1!)(p-2)! + ... + (p-1)!(0!)

               = -1 (mod p)

               = |X| |Y| cos(alpha)

where alpha is the angle between X and Y.  Since X and Y have the same
components in reverse order, we know that |X| = |Y|, so we have  

    [ (0!)^2 + (1!)^2 + ... + ((p-1)!)^2 ] cos(alpha) =  -1  (mod p)

This relates the sum of the squares of the factorials 0! to (p-1)! with
the cosine of the angle between X and Y.  Of course, this doesn't prove 
that the sum of squares is not divisible by p, it just rephrases the
question in terms of the cosine of the angle between two vectors.

Anyway, for any positive integer n let ![n,k] denote the sum of the 
kth powers of the factorials from 0! to (n-1)!.  For example, the 
preceding formula can be written as ![p,2] cos(alpha) = -1 (mod p).  

By Fermat's Little Theorem, the sequence of integers ![p,k] (mod p) 
for k=0,1,2,... is periodic with a period dividing p-1.  Therefore, 
we have
                    ![p,(p-1)m] = 0   (mod p)

for any natural number m.  In general, we have ![p,k]=0 (mod p) if 
and only k is of the form  (p-1)m + c  where c is one of a finite 
set of values for any given prime p.  The question of whether !p 
is ever divisibly by p is equivalent to asking if c can ever be 1.

The values of c for the first few primes are tabulated below:

   p    c values        p     c values        p        c values
 ----  ----------     ----  -----------      ----  ---------------------
   2   0               53   0                127   0  57
   3   0               59   0  53            131   0 105
   5   0  3            61   0  25            137   0  55
   7   0               67   0                139   0
  11   0               71   0                149   0   4  23  26 122 144
  13   0 11            73   0  10  62        151   0
  17   0  7            79   0                157   0
  19   0               83   0  61            163   0
  23   0  6 16         89   0  33            167   0
  29   0  9 23         97   0                173   0
  31   0              101   0                179   0
  37   0 35           103   0  44  58        181   0   6 163 174
  41   0              107   0                191   0
  43   0              109   0  18  57  90    193   0 139
  47   0  9 15        113   0  50  62        197   0

Some additional conjectures arise from the following table: 

      k       values of p^n such that ![p^n,k] = 0 (mod p^n)
    -----    -----------------------------------------------
      1       2
      2       2, 3 
      3       2, 5, 25
      4       2, 3, 5, 9, 149
      5       2
      6       2, 3, 7, 23, 181, 443
      7       2, 5, 17
      8       2, 3, 5, 367
      9       2, 29, 47
     10       2, 3, 11, 73
     11       2, 5, 13
     12       2, 3, 5, 7, 13

Note that if a1, a2,... are the highest powers of the primes p1, 
p2,... such that (pj)^(aj) divides ![(pj)^(aj),k], then the integers 
N that divide ![N,k] are all those of the form  

                  N = (p1)^(b1) (p2)^(b2)... 

with bj <= aj.

Hare-brained Conjectures:
 1.  ![n,1] is not divisible by n for any n>2. (This is the original)
 2.  ![n,5] is not divisible by n for any n>2. (verified for p<1000)
 3.  For any k the number of primes p that divide !(p,k) is finite.
 4.  There is at least one odd prime p that divides !(p,k) for each
     k except k=1 and k=5.