A question about an exercise on basic cardinal arithmetic.

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I just want to make sure that I have proved the following exercise correctly.

Given two cardinal numbers $a$ and $b$ where $a$ is infinite. I was to show that $2\le b \le 2^a \implies b^a=2^a$

I will show that $2^a\le b^a\le 2^a$.

Since $2\le b$, we have an injecton $\sigma:\{1,2\}\rightarrow B$.

Letting $H(A,B)$ be defined as the set of all the functions from $A$ to $B$.

Define $\lambda:H(A,\{1,2\})\rightarrow H(A,B), \lambda(f)=\sigma \cdot f$

this is of course an injection and we have $2^a\le B^a$

Similarly we can show that $b^a\le (2^a)^a=2^a$.

Does this look right?

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The proof is correct. At the end you might want to say explicitly that you invoke the Cantor-Bernstein theorem to conclude equality.