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If $1 \leq r \leq n,\;$ then there are $\quad\dfrac{n(n-1)\cdots(n-r+1)}{r}\quad r$-cycles in $S_n$.

I think that I used induction to prove this, but I don't know how to prove this. Thanks in advance!

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    Define all of your terms, and show all the work you have tried so far! – 2017-02-16
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    Your second sentence is confusing: "I think that I used induction to prove this...."... "...but I don't know how to prove this." That's a claim that you have proven it, but don't know how to prove it? – 2017-02-16
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    Nice find, @DietrichBurde Unfortunately, I already voted to close as "off topic / lack of context"... and if I hadn't, I would be able to use the "duplicate hammer" to unilaterally close it! Argh! – 2017-02-16
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    @amWhy You have another case [here](http://math.stackexchange.com/questions/2147710/counting-the-number-of-m-cycles-in-s-n) now. – 2017-02-16
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    ${ \text{Hello!}}...{\small \text{hello!}}...{\tiny \text{hello!}}$...Is George $\;\;{\text{out there}} .. {\small \text{out there}} ... {\tiny \text{out there!}}$ – 2017-02-16
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    @amWhy Wait! Whose are you closing? Not mine, I hope. I even explained in my post that my approach is slightly different. Mine depends on the fact that $m$ -cycle is simply a bijection $σ:\{1,...,n\}→\{1,...,n\}$ such that $σ$ over $\{1,...,n\}∖D$ is the identity function, but is not the identity function over $D$; i.e., the integers in $D$ appear $σ$'s cycle decomposition, while those in $\{1,...,n\}∖D$ do not., while most others go about finding the number of ways to construct an $m$-cycle. – 2017-02-16
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    @user193319 Okay...I reopened your question...I'm in a good mood today! :-) – 2017-02-16
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    @amWhy, about my comment that I didn't know how to solve the question, I expected a hint, but the other topic answer my question, sorry for the duplicate topic, but I didn't find a similar question when I searched. – 2017-02-16

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I think it's easier to count them. If you have a cycle of lenght $r$, then think it as a list of $r$ ordered numbers. Then, see that you can have $n(n-1)\dots (n-r+1)$ ordered lists of $r$ numbers. After that, notice that you want to find cycles, you will have that if you rotate the ordered list, you will have a different ordered list, but will represent the same cycle. For example, if $r=3$, the rotated ordered lists for $(1,2,3)$ are

$$(1,2,3),(2,3,1),(3,1,2)$$

So, since you have $r$ ordered lists that represent the same cycle (because you can rotate $r$ times the ordered list before getting the first you take to rotate), you will have to divide by $r$ the number of ordered lists and get the number of cycles of lenght $r$.