I have a question concerning infinite cardinals which I found on an old exam paper:
Let $c=2^{\aleph_0}$, $x=2^c, y=2^{2^c}, z=2^{2^{2^c}}$. Put $x^{y^z}, x^{z^y}, y^{z^x}$ in ascending order.
I'm sure I'm being dense but I can't seem to work out how to do this in the casual manner that the question suggests.
Work them from the top down, noting that $y=2^x$ and $z=2^y$ and hence that $\mathfrak{c}<x<y<z$.
$$\large y^z=\left(2^x\right)^z=2^{x\cdot z}=2^z\;,$$
so
$$\large x^{y^z}=\left(2^{\mathfrak{c}}\right)^{2^z}=2^{\mathfrak{c}\cdot 2^z}=2^{2^z}\;.$$
Next,
$$\large z^y=\left(2^y\right)^y=2^{y^2}=2^y=z\;,$$
so
$$\large x^{z^y}=x^z=\left(2^\mathfrak{c}\right)^z=2^{\mathfrak{c}\cdot z}=2^z\;.$$
And finally,
$$\large z^x=\left(2^y\right)^x=2^{y\cdot x}=2^y=z\;,$$
so
$$\large y^{z^x}=y^z=\left(2^x\right)^z=2^{x\cdot z}=2^z\;.$$
Thus,
$$\large y^{z^x}=x^{z^y}<x^{y^z}\;.$$