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I had this as an assumption in my textbook (in the section Poisson approximation to Binomial distribution) but couldn't prove it. textbook - image for reference

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  • $\begingroup$ $n! /(n-k)! \geq (n-k+1)^k$ $\endgroup$ – kingW3 Jun 7 at 16:39
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You are trying to prove that $$\frac{n(n-1)...(n-k+1)}{n^k}\to 1$$ as $n\to \infty$. Note that the product above has a finite number (k) of terms, and so when finding the limit of the above product, you may look at the product of the limits: $$\lim_{n\to\infty}\frac{n(n-1)...(n-k+1)}{\underbrace{n\cdot n\cdot ...\cdot n}_{k}}\\=\bigg(\lim_{n\to\infty}\frac{n}{n}\bigg)\bigg(\lim_{n\to\infty}\frac{n-1}{n}\bigg)...\bigg(\lim_{n\to\infty}\frac{n-k+1}{n}\bigg)$$ Each of these limits equals $1$, so the limit is $1$.

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