Stirling numbers of the first kind special values proof

Prove that

$$\left[n\atop 1\right] = \Gamma(n)$$

Use the recurrence relation

$$\left[n+1\atop k\right] = n\left[n\atop k\right] + \left[n\atop k-1\right]$$

This implies that for \( k=1 \)

$$\left[n+1\atop 1\right] = n\left[n\atop 1\right] + \left[n\atop0\right]$$

Now use that \( \left[n\atop 0\right] = 0 \) to obtain

$$\left[n+1\atop 1\right] = n\left[n\atop 1\right] $$

This implies that

$$f(n+1) = nf(n)$$

By solving this recurrence relation using that \( f(2) = 1 \)

$$f(n+1) = \Gamma(n+1) \,\,\, ;n>0$$

 

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