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.

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
