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Question Number 141633 by Willson last updated on 21/May/21

$$1<\mathrm{a}<\mathrm{b},\mathrm{prove}\mathrm{that}:$$ $$b^n=\underset{k=0}{\sum ^n}{(-1)}^k{\mathrm{C}}_n^k{a}^{\frac{ln\left(\underset{p=0}{\sum ^{n-k}}{\mathrm{C}}_{n-k}^p{a}^{n-p}{b}^p\right)}{ln\left(a\right)}}$$

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