study the convergence of $\sum_{n=1}^\infty \log \frac{n+1}{n}$

In the comments, @Marvis shows how to clean up your proof. Here's another approach. Let $f_n = \log \frac{n+1}{n}$. Examine the ratio of successive terms for large $n$, $$\frac{f_{n+1}}{f_{n}} = 1 - \frac{1}{n} + O\left(\frac{1}{n^2}\right).$$ Therefore, the series diverges by Gauss's test.

Your procedure is fine. I think it easiest to remove the logarithm immediately:

$e^{a_k} = \prod_{n=1}^k{\frac{n+1}{n}} = k+1$

So, $a_k = \log{(k+1)} \rightarrow \infty$.

Your procedure is correct. If you want to write out things more clearly I suggest that you write down the $n$th partial sums $$\begin{align} s_n &= \sum_{i=1}^{n} \log\left(\frac{i+1}{i}\right) \\ &= \sum_{i=1}^{n} \log(i+1) - \log(i) \\ &= [\log(2) - \log(1)] + \dots [\log(n+1) - \log(n)] \\ &= \log(n+1). \end{align} $$ Hence $$ \lim_{n \to \infty} s_n = \lim_{n\to \infty} \log(n+1) = \infty.$$ So then you say that since the limit does not exist, the series is divergent by definition.

Note: The notation is important. It is not correct to write $\lim_{n\to \infty} \log(n+1) \to \infty$, we write $\lim_{n\to \infty} \log(n+1) = \infty.$

Your procedure is correct. I suggest another way:

$\log \frac{n+1}{n}$ is positive, and: $\log \left(1+\frac{1}{n}\right)\sim \frac{1}{n}$ for $n \rightarrow\infty$. So, we have:

$\sum \frac{1}{n}$ that diverges.