I want to check if the following statement holds:
If $a, b, c $ are cardinal numbers and $a<b$ then $a^c<b^c$.
If the statement would hold, given that there is a bijective funcion $f: a \to b$ there will also be a bijective function $g: a^c \to b^c$.
Since $f$ is bijective, we have that $f(a)=b$.
So there will be a $y$ such that $f(a^c)=y$.
Is this right so far? If so, how can we continue?
Your statement is false. For example, $\mathbb{N}^\mathbb{N}$ and $2^\mathbb{N}$ have the same cardinality - put another way, $$(\aleph_0)^{\aleph_0}=2^{\aleph_0}.$$
We can see this by invoking basic cardinal arithmetic rules, as $$2^{\aleph_0}\le (\aleph_0)^{\aleph_0}\le (2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\times\aleph_0}=2^{\aleph_0},$$ with the first inequality being trivial, the second following from the weak monotonicity of exponentiation (which is also pretty trivial), and the third and fourth following from the definitions of cardinal multiplication and exponentiation (together with a bit of Currying).
However, we can also see this directly via Cantor-Bernstein (which, as a reminder, does not require Choice). Namely, fix your favorite pairing function $\langle\cdot,\cdot\rangle:\mathbb{N}^2\rightarrow\mathbb{N}$, and given $f:\mathbb{N}\rightarrow\mathbb{N}$ let $\hat{f}:\mathbb{N}\rightarrow 2$ be defined by $$\hat{f}(\langle a,b\rangle)=1\quad\iff\quad f(a)=b$$ (and $\hat{f}(\langle a,b\rangle)=0$ otherwise). It's easy to check that the map $$f\mapsto\hat{f}$$ is an injection from $\mathbb{N}^\mathbb{N}$ to $2^\mathbb{N}$. Since the identity provides an injection the other way, Cantor-Bernstein gives a bijection.
(Incidentally, your "proof" also gets cardinal inequality wrong: "$a<b$" means that there is an injection from $a$ to $b$ but there is not an injection from $b$ to $a$.)
It's worth pointing out that increasing the exponent doesn't necessarily increase the whole power either: letting $\kappa=2^{\aleph_1}$, we have $$\kappa^{\aleph_0}=2^{\aleph_1\times\aleph_0}=2^{\aleph_1}\quad\mbox{and}\quad \kappa^{\aleph_1}=2^{\aleph_1\times\aleph_1}=2^{\aleph_1}.$$ This is indeed the usual situation: