Solve recurrence relation $T(n)=n^{1/5}T(n^{4/5})+5n/4$
Derivation
Lets consider $n = 2^{(5/4)^k}$. Now we can evaluate step by step:
$$ S(n) = S(2^{(5/4)^{k-1}}) + 5/4 = S(2^{(5/4)^{k-2}}) + 5/4 + 5/4$$
Notice that every time $k$ is reduced by one since:
$$\left(2^{(5/4)^{k}}\right)^{4/5} = 2^{(5/4)^{k}\cdot(4/5)} = 2^{(5/4)^{k-1}}$$
Thus we can denote $S(n)$ as (assuming $S(2) = 0$ as base case):
$$\sum_{i=1}^k 5/4 = (5/4)k $$
Now lets solve $n = 2^{(5/4)^k}$ for $n$. Apply $\log_2$ and then $\log_{5/4}$ on both sides: $k = \log_{5/4}(\log_2(n))$ and thus $S(n) = (5/4)\log_{5/4}(\log_2(n))$.

Test by plugging it in
You can simply plug your solution in to test it.
For $S(n) = S(n^{4/5}) + 5/4$ you can obtain $S(n) = (5/4)\log_{5/4}(\log_2(n))$ by the recursion tree method (as described above). Lets plug this into our recurrence:
$$S(n) = (5/4)\log_{5/4}(\log_2(n^{4/5}))+5/4 = (5/4)\log_{5/4}((4/5)\log_2(n))+5/4$$
And by the rules of logaritm:
$$S(n) = (5/4)\log_{5/4}(\log_2(n))+(5/4)\log_{5/4}(4/5)+5/4$$
Notice $\log_{5/4}(4/5) = -1$ and things cancel out:
$$S(n) = (5/4)\log_{5/4}(\log_2(n))-(5/4)+5/4 = (5/4)\log_{5/4}(\log_2(n))$$
Our expression for $S(n)$ fulfills the recurrence. Hence $$T(n)/n = S(n) \Rightarrow T(n) = (5/4)n\log_{5/4}\log_2(n)$$
A double logarithm looks scary but if often occurs when recurrences have the form $T(n) = aT(n^\alpha) +b$.