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)?

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