Simple/efficient representation of Stirling numbers of the first kind Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum
$$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$
This can be used for direct calculation of $S_2(n,k)$, without the need to compute any preceding values. But for Stirling numbers of the first kind, one seems to need a nested sum or a recurrence over preceding values, the most direct known representation perhaps being
$$S_1(n,k) = \sum_{j=0}^{n-k} (-1)^j {n+j-1\choose n-k+j} {2n-k \choose n-k-j} S_2(n-k+j,j). \qquad (2)$$
Is there a reason to believe that no formula similar to (1) exists for Stirling numbers of the first kind? Does a formula better than (2)+(1) for calculations exist (assume that I have no interest in generating a table of all preceding values)?
 A: Would you, or would you not, consider as "simple" integral and/or series representations that work for complex values, suitably restricted?
A: $S_1(n,0)=S_1(0,n)= \delta_n \; \;$ and, for $n > 0$, 
$$S_1(n,k)= \lim_{y \to 0} \;  \frac{y^{-k}}{k!} \; \sum_{j=1}^k (-1)^j \binom{k}{j} \; \frac{(-j \; y)!}{(-j \; y-n)!} \;$$
$$ = \lim_{y \to 0} \;  \frac{y^{-k}}{k!} \; \sum_{j=1}^k (-1)^{n-j} \binom{k}{j} \; \frac{(j \; y - 1 + n)!}{(j \; y-1)!} \; $$
$$= \sum_{j=k}^n \; S_1(n,j)\; (-y)^{j-k}\;S_2(j,k) \; |_{y=0} \; . $$
For a derivation, see A class of differential operators and the Stirling numbers. Note that with $y$ small enough taking the nearest integer generates $S_1$.
A: You ask "Is there a reason to believe that no formula similar to (1) exists for Stirling numbers of the first kind?" One reason to believe that there is no such formula is that Louis Comtet (Advanced Combinatorics, p. 216) says so: "... the Stirling number of the second kind s(n, k) can be expressed as a single summation of elementary terms, that is, which are themselves products and quotients of factorials and powers. There does not exist an analogous formula for the numbers of the first kind, the ‘shortest formula’ [7a, a’] below being a double summation of elementary terms." 
A proof of Comtet's nonexistence assertion would be interesting.
A: If Stirling numbers of the first kind are the numbers associated with the Stirling series, if there is a "sufficiently simple-to-compute" representation of them, you can factor integers in time polynomial in the number of their bits, using a simple property presented in a blog post by Richard Lipton and a particular rational/exponential approximation to $n!$ that's based on the Stirling series.  I spent some time looking for such a representation once, without any luck, though.
It's believed by many that there is no such algorithm to factor integers, (although Richard has written several posts suggesting that it's still rather uncertain), so if they're right, there is no "sufficiently simple-to-compute" representation of the Stirling numbers of the first kind.
A: http://members.lycos.co.uk/sobalian/index.html 
OEIS A008275 
a(n,k) = s(k,n) = (-1)^(k-n) * S1(k,n) = ( (-1)^(k-n) ) * ( k!/{(n-1)!*2^(k-n)} 
               ) * [ { 1/(k-n)! }k^(k-n-1) - { (1/6)(1/(k-n-2)!) }k^(k-n-2) + 
               { (1/72)(1/(k-n-4)!) }k^(k-n-3) - { (1/6480)(5/(k-n-6)! -36/(k-n-4)!) 
               }k^(k-n-4) + { (1/155520)(5/(k-n-8)!-144/(k-n-6)!) }k^(k-n-5) 
               - { (1/6531840)(7/(k-n-10)! -504/(k-n-8)!+2304/(k-n-6)!) }k^(k-n-6) 
               + { (1/1175731200)(35/(k-n-12)!-5040/(k-n-10)!+87264/(k-n-8)!) }k^(k-n-7) 
               - { (1/7054387200)(5/(k-n-14)!-1260/(k-n-12)!+52704/(k-n-10)!-186624/
               (k-n-8)!) }k^(k-n-8) + { (1/338610585600)(5/(k-n-16)!-2016/(k-n-14)!+164736/
               (k-n-12)!-2156544/(k-n-10)!) }*k^(k-n-9) - ..... ] 
