Using standard asymptotic methods,

    Gamma(x+1,x)

       |\oo  -t  x
    =  |    e   t  dt
      \| x

       x  -x    |\oo  -x(s-log(s+1))
    = x  e   x  |    e               ds     [t = x(s+1)]
               \| 0

       x  -x    |\oo  -xu^2/2
    = x  e   x  |    e        s' du     [u^2/2 = s-log(s+1)] [1]
               \| 0

Compare this to

    Gamma(x+1)

       |\oo  -t  x
    =  |    e   t  dt
      \| 0

       x  -x    |\oo  -x(s-log(s+1))
    = x  e   x  |    e               ds     [t = x(s+1)]
               \| -1

       x  -x    |\oo  -xu^2/2
    = x  e   x  |    e        s' du     [u^2/2 = s-log(s+1)] [2]
               \|-oo

Working from ss' = (1+s)u in a fairly straightforward manner, we get

        oo
        ---     k
    s = >   a  u                                             [3]
        ---  k
        k=0

where a_0 = 0, a_1 = 1, and for k > 1,

                      k-2
          1         1 ---
    a  = --- a    - - >   a    a                             [4]
     k   k+1  k-1   2 ---  k-j  j+1
                      j=1

Recursion [4] gives that

            1  2    1  3    1   4     1   5     1    6
    s = u + - u  + -- u  - --- u  + ---- u  + ----- u  + ...
            3      36      270      4320      17010

therefore,

             2      1  2    2   3    1   4     1   5
    s' = 1 + - u + -- u  - --- u  + --- u  + ---- u  + ...   [5]
             3     12      135      864      2835

In the asymptotic expansion of [2], the odd order terms of s' get
eliminated; however, in the asymptotic expansion of [1], they do not.
Thus, to get the terms in the asymptotic expansion of [2], we would
throw out the odd order terms in s' and double the even order terms.

To aid in using [1] with [5], we compute

     |\oo  -bu^2  k      1  |\oo  -bu^2  k-1
     |    e      u  du = -  |    e      u    d(u^2)
    \| 0                 2 \| 0

                         1  |\oo  -bu  (k-1)/2
                       = -  |    e    u        du
                         2 \| 0

                         1  -(k+1)/2       k+1
                       = - b         Gamma(---)              [6]
                         2                  2

Putting together [1], [5], and [6], we get

    Gamma(x+1,x)

       |\oo  -t  x
    =  |    e   t  dt
      \| x

       x  -x x        1/2              2      1
    ~ x  e   - [ (2/x)    Gamma(1/2) + - (2/x)  Gamma(1)
             2                         3

       1      3/2               2       2
    + -- (2/x)    Gamma(3/2) - --- (2/x)  Gamma(2)
      12                       135

       1       5/2                1       3
    + --- (2/x)    Gamma(5/2) + ---- (2/x)  Gamma(3) ]
      864                       2835

       x  -x   1                2    1
    = x  e   [ - sqrt(2 pi x) + - + --- sqrt(2 pi x)
               2                3   24x

        4      1                      8    
    - ---- + ------ sqrt(2 pi x) + ------- ]
      135x   576x^2                2835x^2

       x  -x                1    1      1
    = x  e   sqrt(2 pi x) [ - + --- + ------ + ... ]
                            2   24x   576x^2

       x  -x   2     4       8
    + x  e   [ - - ---- + ------- + ... ]                    [7]
               3   135x   2835x^2

We have collected the first half of [7] to contain those terms that come
from the even order terms of s' and the second half to contain the terms
that come from the odd order terms of s'.  By tossing out the terms that
come from the odd order terms of s' and doubling, we get, as mentioned
earlier, Stirling's Formula:

    Gamma(x+1)

       |\oo  -t  x
    =  |    e   t  dt
      \| 0

       x  -x                     1      1
    ~ x  e   sqrt(2 pi x) [ 1 + --- + ------ + ... ]         [8]
                                12x   288x^2

Combining [7] and [8] we get that the ratio of

    Gamma(x+1,x)/Gamma(x+1)

      1        1         2    23       23
    ~ - + ------------ [ - - ---- + ------- + ... ]          [9]
      2   sqrt(2 pi x)   3   270x   3024x^2

Rob Johnson