Is $0^\infty$ indeterminate?
No, it is zero.
Consider the function $f(x,y) = x^y$ and consider any sequences $\{(x_0, y_0), (x_1, y_1), \ldots\}$ with $x_i \to 0$ and $y_i \to \infty$. It is easy to see that $f(x_n,y_n)$ converges to zero: let $\epsilon > 0$. For some $N$, $|x_i| < \epsilon$ and $y_i > 1$ for all $i \geq N$, so $|f(x_i,y_i)| < \epsilon$ for all $i\geq N$.
More generally, as $x\to c$ and $y\to \infty$, $x^y$ converges to 0 for $|c|<1$, diverges to infinity for $c>1$, oscillates without converging for $c \leq -1$, and is indeterminate when $c=1$.
Since the question came to the front page again, let's do a complex example, to show that (at least this time) Mathematica is not crazy...
For $t>0$, let $f(t) = t, g(t) = i/t$. Then
$$
\lim_{t\to 0+} f(t) = 0,\qquad \lim_{t\to 0+}g(t) = \infty
\\
f(t)^{g(t)} = \exp\frac{i\;\log t}{t}
$$
and $\mathrm{Re} \Big(f(t)^{g(t)}\Big)$ (pictured) does not converge as $t \to 0^+$.
