# Proving $\prod_{i=1}^n (\frac{1}{i} + 1) = n+1$

Prove using a direct proof that

$$\prod_{i=1}^n \left(\frac{1}{i} + 1\right) = n+1$$

Okay, so I think I have done it correctly using an inductive proof:

Base case: $(1+\frac11)=2$, therefore $n=1$ is true. Assume that $n=k$ is true for some $k \geq 1$. Now we must prove that $n=k+1$ is true.

$$\left(1+\frac11\right)\left(1+\frac12\right)\cdots\left(1+\frac1k\right)\left(1+\frac1{k+1}\right) = (k+1)+1 = k+2$$

therefore $n=k+1$ is true and QED by mathematical induction. But by direct proof I am sort of lost.

To relate each term with their previous one, I came to this step.

$$1+\frac{1}{k-1} = 1+\frac{1}{k} - \frac{1}{k-k^2}$$

But I'm quite confused at this stage anyway. Any help or advice would be very appreciated.

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We have $$\prod_{i=1}^n \left(\frac{1}{i} + 1\right) = \prod_{i=1}^n \frac{i+1}{i} = \frac{\prod_{i=1}^n (i+1)}{\prod_{i=1}^n i} = \frac{(n+1)!}{n!} = n+1.$$
$$\prod_{i=1}^n\left(\frac1i+1\right)=\prod_{i=1}^n\left(\frac{1+i}i\right)=\frac{\displaystyle\prod_{i=1}^n(1+i)}{\displaystyle\prod_{i=1}^ni}=\frac{\displaystyle\prod_{i=2}^{n+1}i}{\displaystyle\prod_{i=1}^ni}=n+1$$