Show that the union of two sets of the same cardinality has again the same cardinality. Greetings great wise ones.
Continuing my set-theoretic adventures, I have again stumbled upon a problem and need guidance.
The original problem goes like this:
Let $k_0 = \aleph_0$ and for any $n > 0$, $k_{n+1} = 2^{k_n}$ (These are Beth numbers basically).
I need to show that $k_n + k_n = k_n$ for any $n$.
So what I'm trying to do so far, is conjure up a bijection $f : A \rightarrow A \cup B$ where $A$ and $B$ are two disjoint sets whose cardinality is $k_n$. An injection from $A$ to $A \cup B$ is easy, the tricky part is other way around, but I am not sure how to go around doing that.
Trying to find some help in a bunch of set theory books I have, I found that I could prove the aforementioned equality by first proving that $k_n * k_n = k_n$ (in that $k_n + k_n = (1 + 1)k_n = 2k_n = k_n$), but that problem is pretty much the same as what I'm having trouble with.
That book discussed cardinals in general though and not ones of the form I gave, so maybe I'm not using the definition of those $k_n$s to my advantage.
Can someone please give me a hint/direction?
 A: First, a little notation: for sets $X, Y$, define their disjoint union to be  $X \sqcup Y := (\{0\}\times X) \cup (\{1\}\times Y)$. Thus for cardinals $\kappa, \lambda$, $\kappa + \lambda = card(\kappa \sqcup \lambda)$. Also define $^X Y = $ the set of all functions $X\to Y$.
For $n=0$, we know an injection $k_0 \sqcup k_0 \to k_0$, namely $(i,n) \mapsto 2n + i$. This maps $\{0\}\times k_0$ to the evens and $\{1\}\times k_0$ to the odds.
Now we can define an injection $k_{n+1} \sqcup k_{n+1} \to k_{n+1}$, which is to say, an injection $$
\{0\}\times\, ^{k_{n}}\!2 \,\cup\, \{1\}\times\, ^{k_{n}}\!2 \,\longrightarrow\, ^{k_{n}}\!2.
$$
For any $f\in {^{k_n} 2}$, $i < 2$, let $i*f \in {^{k_{n}}2}$ be the function
$$
(i*f)(\alpha) = \begin{cases}
i &\text{ if $\alpha = 0$,} \\
f(\alpha + 1)&\text{ if $0 < \alpha<\omega$ (i.e. $\alpha$ is a nonzero integer)}, \\
f(\alpha) &\text{ otherwise.}
\end{cases}$$
Such an $f$ is a binary sequence of length $k_n$, so  $i*f$ just "prepends $i$". Then
$$
(i, f)\mapsto i*f\colon \{0\}\times\, ^{k_{n}}\!2 \,\cup\, \{1\}\times\, ^{k_{n}}\!2 \,\longrightarrow\, ^{k_{n}}\!2
$$
does the trick.
