Answer: For the statement

\(H_n: \sum_{r=1}^{n} r\ln\left(\frac{r+1}{r}\right)=\ln\left(\frac{(n+1)^n}{n!}\right)\),

we prove \(H_n\) by induction.

When \(n=1\),

\(\sum_{r=1}^{1} r\ln\left(\frac{r+1}{r}\right)=1\cdot \ln 2=\ln 2=\ln\left(\frac{2^1}{1!}\right)\),

so \(H_1\) is true.

Assume that for some positive integer \(k\),

\(\sum_{r=1}^{k} r\ln\left(\frac{r+1}{r}\right)=\ln\left(\frac{(k+1)^k}{k!}\right)\).

Then

\(\sum_{r=1}^{k+1} r\ln\left(\frac{r+1}{r}\right)=\sum_{r=1}^{k} r\ln\left(\frac{r+1}{r}\right)+(k+1)\ln\left(\frac{k+2}{k+1}\right)\)

\(=\ln\left(\frac{(k+1)^k}{k!}\right)+(k+1)\ln\left(\frac{k+2}{k+1}\right)\)

\(=\ln\left(\frac{(k+1)^k}{k!}\right)+\ln\left(\left(\frac{k+2}{k+1}\right)^{k+1}\right)\)

\(=\ln\left(\frac{(k+1)^k (k+2)^{k+1}}{k!(k+1)^{k+1}}\right)\)

\(=\ln\left(\frac{(k+2)^{k+1}}{(k+1)!}\right)\).

Thus \(H_k \Rightarrow H_{k+1}\). Since \(H_1\) is true, it follows by mathematical induction that

\(\sum_{r=1}^{n} r\ln\left(\frac{r+1}{r}\right)=\ln\left(\frac{(n+1)^n}{n!}\right)\)

for all positive integers \(n\).

Let

\(H_n: \sum_{r=1}^{n} r\ln\left(\frac{r+1}{r}\right)=\ln\left(\frac{(n+1)^n}{n!}\right)\).

We prove \(H_n\) by mathematical induction.

Base case: When \(n=1\),

\(\sum_{r=1}^{1} r\ln\left(\frac{r+1}{r}\right)=1\cdot \ln\left(\frac21\right)=\ln 2\).

Also,

\(\ln\left(\frac{(1+1)^1}{1!}\right)=\ln\left(\frac{2^1}{1!}\right)=\ln 2\).

So \(H_1\) is true.

Inductive step: Assume that for some positive integer \(k\),

\(\sum_{r=1}^{k} r\ln\left(\frac{r+1}{r}\right)=\ln\left(\frac{(k+1)^k}{k!}\right)\).

We must show that \(H_{k+1}\) is true. Starting from the left-hand side for \(k+1\),

\(\sum_{r=1}^{k+1} r\ln\left(\frac{r+1}{r}\right)=\sum_{r=1}^{k} r\ln\left(\frac{r+1}{r}\right)+(k+1)\ln\left(\frac{k+2}{k+1}\right)\).

Using the induction hypothesis,

\(=\ln\left(\frac{(k+1)^k}{k!}\right)+(k+1)\ln\left(\frac{k+2}{k+1}\right)\).

Bring the factor \(k+1\) into the logarithm:

\(=\ln\left(\frac{(k+1)^k}{k!}\right)+\ln\left(\left(\frac{k+2}{k+1}\right)^{k+1}\right)\).

Combine the logarithms:

\(=\ln\left(\frac{(k+1)^k (k+2)^{k+1}}{k!(k+1)^{k+1}}\right)\).

Now simplify the fraction. Since \((k+1)!=(k+1)k!\), we have

\(k!(k+1)^{k+1}=(k+1)!\,(k+1)^k\).

Therefore

\(\ln\left(\frac{(k+1)^k (k+2)^{k+1}}{k!(k+1)^{k+1}}\right)=\ln\left(\frac{(k+2)^{k+1}}{(k+1)!}\right)\).

This is exactly the statement \(H_{k+1}\).

Hence \(H_k \Rightarrow H_{k+1}\), and since \(H_1\) is true, it follows by induction that

\(\sum_{r=1}^{n} r\ln\left(\frac{r+1}{r}\right)=\ln\left(\frac{(n+1)^n}{n!}\right)\)

for all positive integers \(n\).