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Question Number 188798 by mathlove last updated on 07/Mar/23

Answered by ARUNG_Brandon_MBU last updated on 07/Mar/23

$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{2^x{3}^x{4}^x{5}^x...n^x-1}{x}$$$$=\underset{x\to 0}{\lim }\frac{A^x-1}{x},[A=n!]$$$$=\ln A=\ln (n!)$$

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