Prove $a \in\mathbb I, b \in \mathbb I \implies a^b \in \mathbb I$ Let $\mathbb I$ be the set of Irrational numbers and  $\mathbb Q$ be the set of Rational numbers. Show $a \in \mathbb I, b \in\mathbb I\implies a^b \in\mathbb I$

Using a counter example I can prove this:

$a = \sqrt{2}^\sqrt{2}$ 
$b = \sqrt{2}$ 

$\sqrt{2}^{\sqrt{2}\times\sqrt{2}}  = 2$ 

since $Q \subset I$ we can say $2 \in I$

is this correct? Or is there a rigorous proof? Thanks. :)
Your proof is a sketch of the proof that there are irrational numbers $a,\,b$ such that $a^b$ is rational - that is, assuming that you accept the law of excluded middle (that a proposition must be either true or false) so this is not a proof to an intuitionist: but I digress.

You argue as follows. Let $a = \sqrt{2}^\sqrt{2}$. Now either $a$ is rational or it is not (this is where an intuitionist would reject the proof).

If it is irrational, then raising to the irrational power $\sqrt{2}$ yields a rational number, to wit $2$. Therefore we have an example of an irrational ($a$) to an irrational power that is rational.

On the other hand, if $a$ is not irrational, it is rational, and so $a$ itself stands as an example of an irrational ($\sqrt{2}$) raised to a irrational power ($\sqrt{2}$) which yields a rational result.

Either way, we have a counterexample. Curiously, there is no way from this proof to tell which one is the counterexample: we simply know we have one, disproving the proposition.
To show that
$a^b \in I$,
you have to use induction.

Assumptions:
$I$ means the natural (positive) integers,
$a \in I$ and $b \in I$.

Needed results:
Inductive definition of exponentiation
and the product of
two elements of $I$
is also in $I$.

Base step:
For $b=1$,
$a^1 = a \in I$.

Suppose true for $b$
(i.e., $a^b \in I$).
Then

$\begin{array}\\
a^{b+1}
&amp;=a^b\cdot a
\quad\text{(by the inductive definition of exponentiation)}\\
&amp;\in I
\quad\text{(because $a^b \in I, a \in I$, and product of elements in $I$ is in $I$)}\\
\end{array}
$