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Firstly, how would you solve an equation such as the Binomial Theorem by way of Mathematical Induction, and how could you use that to prove the following?

$$\left(1 + \frac1n\right)^n = 1 +\sum_{k=1}^n \left[ \frac1{k!}\prod_{r=0}^{k-1}\left(1 - \frac{r}n\right)\right]$$

You may recognize this question from Apostol's Calculus book.

So you know, I am not asking because I want help with a homework problem. Rather, I am 16 years old and endeavoring to teach myself the mathematics required for Machine Learning, and subsequently, don't have access to a professor whom I can ask.

Thank you.

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  • $\begingroup$ Maybe it is easier to prove the binomial theorem $(1+x)^{n}=1+\sum_{k=1}^{n} \frac{n!}{k!(n-k)!}x^{k}$ by using induction (this is not hard) and put $x=1/n$. $\endgroup$
    – Seewoo Lee
    Commented Apr 13, 2018 at 5:09

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For $n\geq k\geq 1$ we have$$\binom {n}{k}=\frac {n!}{k!(n-k)!} =\frac {1}{k!}\prod_{r=0}^{k-1} (n-r).$$ Therefore for $n\geq k\geq 1$ we have $$n^{-k}\binom {n}{k}=\frac {1}{k!}\cdot n^{-k}\prod_{r=0}^{k-1}(n-r)=$$ $$=\frac {1}{k!}\prod_{r=0}^{k-1}\frac {(n-r)}{n}=$$ $$=\frac {1}{k!}\prod_{r=0}^{k-1}(1-\frac {r}{n})$$ By the Binomial Theorem, if $n\geq 1$ then $(1+n^{-1})^n=1+\sum_{k=1}^{n}n^{-k}\binom {n}{k}$. By the calculations above, this is equal to the RHS formula in your Q.

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