Show that $\prod_{k=1}^n \left(\prod_{j=1}^k\frac{k}{j}\right)$ is always an integer.

Note that $$\prod_{k=1}^{n} \binom{n}{k}=\prod_{k=1}^{n} \frac{n^{\underline{k}}}{k!}=\prod_{k=1}^n \frac{k^k}{k!}$$ As $$\prod_{k=1}^{n}n^{\underline{k}}=\prod_{k=1}^n k^k$$ This follows elementarily from following the terms.

NOTE:In this context, $n^{\underline{k}}$ signifies the falling factorial function. Be sure not to confuse it with $n^k$.

Your start was good:

$$\prod_{k=1}^n \left(\prod_{j=1}^k\frac{k}{j}\right)$$

First simply convert the inner formula into a closed one, as you did.

$$\prod_{k=1}^n \frac{k^k}{k!}$$

From that point, the things are more easy as they seem. Here is the formula more clearly, without product:

$$\frac{1^1 \cdot 2^2 \cdot 3^3 \cdot ... \cdot n^n}{1! \cdot 2! \cdot ... \cdot n!}$$

We convert also the nominator to a factorial expression:

$1^1 \cdot 2^2 \cdot 3^3 \cdot ... \cdot n^n = \frac{n!}{1!} \cdot \frac{n!}{2!} \cdot ... \cdot \frac{n!}{n!}$

The result is the formula:

$\frac{\frac{n!}{1!} \cdot \frac{n!}{2!} \cdot ... \cdot \frac{n!}{n!}}{1! \cdot 2! \cdot ... \cdot n!}$

From that point, you can easily see that here are the product of the binomial constants:

$\frac{\frac{n!}{1!} \cdot \frac{n!}{2!} \cdot ... \cdot \frac{n!}{n!}}{n! \cdot (n-1)! \cdot ... \cdot 1!}$

$\frac{n!}{1! \cdot n!} \cdot \frac{n!}{2! \cdot (n-1)!} \cdot ... \cdot \frac{n!}{n! \cdot 1!}$

Q.E.D.