Prove That the Following Function Always Evaluates to Integers

Question

The function $f(n)$ is defined over $\mathbb{N}$ as follows

$$f(n) = \prod_{i=1}^{n}i^i$$

Show that $g(n,r)$ where $r \in \mathbb{N}$ and $1 < r < n$ which is defined as

$$g(n,r) = \frac{f(n)}{f(r)\times f(n-r)}$$

takes only integer values.

I realized that it is enough to show that the power of each prime in the numerator is always greater than or equal to the power of each prime in the denominator but was unable to come up with anything.

Any hints and ideas are welcome. Thanks!

Answer 1

Let $m=\max (r,n-r).$

We have $f(n)/f(m)=\prod_{j=m+1}^n j^j$ which is an integer multiple of the integer $\prod_{j=m+1}^nj^m$ which is an integer multiple of the integer $\prod_{j=m+1}^nj^{n-m}$ because $m\geq n-m.$

And $f(n-m)=\prod_{j=1}^{n-m}j^j$ which is an integer divisor of the integer $\prod_{j=1}^{n-m}j^{n-m}.$

Now $\prod_{j=m+1}^nj^{n-m}/\prod_{j=1}^{n-m}j^{n-m}=\binom {n}{n-m}^{n-m}$ is also an integer.