The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$. Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$.
The best way to solve this problem is to get the iterated part out of the exponent:
\[5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1\]
\[5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}\]
\[5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}\]
\[a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}\]
\[a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}\]
Plug in $n = 1, 2, 3, 4$ to see the first few terms of the sequence: \[\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.\] We notice that the terms  $5, 8, 11, 14$ are in arithmetic progression. Since $a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}$, we can easily use induction to show that $a_n = \log_5{(3n + 2)}$. So now we only need to find the next value of $n$ that makes $\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$. We test $25$: \[3n + 2 = 25\] \[3n = 23\] This has no integral solutions, so we try $125$: \[3n + 2 = 125\] \[3n = 123\] \[n = \boxed{041}\]