Prove that $(4/5)^{\frac{4}{5}}$ is irrational Prove that $(4/5)^{\frac{4}{5}}$ is irrational.
My proof so far:
Suppose for contradiction that $(4/5)^{\frac{4}{5}}$ is rational.
Then $(4/5)^{\frac{4}{5}}$=$\dfrac{p}{q}$, where $p$,$q$ are integers.
Then $\dfrac{4^4}{5^4}=\dfrac{p^5}{q^5}$
$\therefore$ $4^4q^5=5^4p^5$
I've got to this point and now I don't know where to go from here.
 A: Hint $ $ By unique factorization, comparing powers of $\,2\,$ as below yields a contradiction $\,5\mid 8$ 
$$\begin{align}2^{\large\color{#c00}8}\, Q^{\large\color{#0a0}5}\! &=\,  P^{\large\color{#f70}5}\, 5^{\large 4}\\ 
\Rightarrow\ \color{#c00}8\! +\! \color{#0a0}5q &=\, \color{#f80}5p\\ 
\Rightarrow\qquad\, 8 &=\, 5(p\!-\!q)\end{align}$$
Remark $\ $ This is a generalization of the classical proof of the irrationality of $\,\sqrt 2\,$ by comparing the parity of powers of $\,2\,$ in $\,P^2\! = 2Q^2.\,$ It compares powers mod $5$ (vs. $2)$ since it involves $5$'th
 vs. $2$nd powers.
The proof does not require the full power of unique prime factorization, only that each natural may be written uniquely in the form $\,2^j n,\,\ n\,$ odd  (which has very a simple inductive proof).
A: Outline: If the number $\alpha$ is rational, there exist integers $p$ and $q$ which are relatively prime such that $\alpha=\frac{p}{q}$.
From your $4^4q^5=5^4p^5$, argue that $5$ divides $q$, and then that $5$ divides $p$. 
A: A slightly modified approach.
$(\frac{4}{5})^\frac{4}{5} = (\frac{4}{5})^{1-\frac{1}{5}} =  (\frac{4}{5})^1 \times(\frac{5}{4})^\frac{1}{5}$
Since the first factor is rational we need only show that the latter term $(\frac{5}{4})^\frac{1}{5}$ is irrational.
Let us assume that the latter is rational and write (where $p$ and $q$ are coprime)
$\frac{p}{q}=(\frac{5}{4})^\frac{1}{5} \implies 4\times p^5 = 5 \times q^5$
The L.H.S is even and this implies that $q$ is even (sub $q=2^5\times n$ in above); $q^5$ then must have a factor of $2^5$. This would then require $p$ to be even and thus violates the coprime assumption and gives the required contradiction.
A: This is essentially the same as the standard proof that $\sqrt 2$ is irrational.
Let $z = (4/5)^{(4/5)}$.  Now to calculate $z^5$:
$$
z^5 = \left(\left({4\over 5}\right)^{4\over 5}\right)^5 =  \left({4\over 5}\right)^{4}
= {4^4\over 5^4}
$$
Clearly $z \neq 0$ as $0^{(5/4)} = 0 \neq 4/5$.
Let a rational number $r = p/q$ where $p$ and $q$ are positive integers.  
Every positive integer is known to have a unique prime factorization.  Specifically, we know there exist integers $i \geq 0$ and $t \geq 1$ such that $p = t \times 5^i$ and $t \not \equiv 0 \mod 5$.
Likewise, $q = u \times 5^j$.  Thus, we have
$$
r = \frac pq = \frac tu \times 5^{(i-j)}
$$
and neither $t$ nor $u$ is divisible by $5$.
Now, we calculate $r^5$:
$$
r^5 = \frac{u^5}{v^5} \times 5^{5 (i-j)}
$$
If $r^5 = z^5$, then we must have $5(i-j) = -4$.  There are clearly no integers $i,j$ which satisfy that.  Thus we have shown that for any rational number $r$ it cannot be the case that $r = z = (4/5)^{(4/5)}$.
QED
