Restrictions for the geometric series formula and why might $0^0 = 1$ I was curious about how the restrictions for the geometric series formula arise from its proof, namely
that $|r|<1$. Here is the traditional proof (for the partial sum of a geometric series) that I'm familar with:
$$\sum_{k=0}^{n-1} ar^n=s=a+ar+ar^2+ar^3+\dots+ar^{n-1}$$
$$\implies rs=ar+ar^2+ar^3+\dots+ar^n$$
$$\implies s-rs=a-ar^n$$
$$\implies s=a\Big(\frac{1-r^n}{1-r}\Big)$$
Immediately we find that since the denominator contains $1-r$, then this formula restricts $r\neq1$. Continuing forward, we find that the infinite sum is simply the limit
$$\sum_{k=0}^\infty ar^n=\lim_{n\to\infty}a\Big(\frac{1-r^n}{1-r}\Big)$$
Using simple tests for convergence, the limit only converges on the restriction that $|r|<1$. Considering $0<1$, there does not seem to be much indication (arising solely from the proof) that $r$ cannot be $0$. Naively so, we could evaluate
$$\sum_{k=0}^\infty 0^k=\lim_{n\to\infty}\Big(\frac{1-0^n}{1-0}\Big)=1$$
Controversially, it seems this can be used as a tool to deduce a value for $0^0$.
$$\implies \sum_{k=0}^\infty 0^n=1=0^0+\sum_{k=1}^\infty0^k$$
$$\implies 0^0=1-\sum_{k=1}^\infty0^k$$
Since $0^k=0\ \ \ \forall k>0$, we find that
$$\sum_{k=1}^\infty0^k=0$$
$$\therefore 0^0=1$$
As we see, on the assumption that the geometric series formula works for all $|r|<1$, then $0^0=1$ follows. I feel that mathematicians are rolling in their grave at this, so

What is the proof missing to show that $r\neq 0$?

Edit: Before making this post, I was under the impression (and education) that $0^0$ was undefined by convention. However it's clear to me now that this seems to be up for debate. Justification for either convention is something I'm very intrigued by.
 A: If you are in a context where your exponents are all integers, I have yet to see a single good reason why $0^0=1$ shouldn't be a convention. And I have seen several reasons why it should be (appealing to the empty product, using elementary combinatorics, and the application to geometric / power series being the three I can think of right now).
As such, setting $0^0$ to be equal to $1$ (at least for integer exponents) solves some issues and as far as I'm aware it raises no issues. It is therefore a good convention. It is also a pretty common convention.
I am partial to letting $0^0=1$ all the time. However, some people think the non-existence of $\lim_{x,y\to0^+}x^y$ (or similar limiting arguments) speaks against this definition in contexts where the exponent is allowed to be any (non-negative) real number. I do not think this is an issue, but to each their own.
A: A useful definition of exponentiation works for non-negative integer, or more generally , if $a,b$ are cardinalities of sets $A,B$, then $a^b$ is the cardinality of the set of functions $B\to A$. With this, $0^0$ quite naturally.
We can extend this definition largely by using $x^y=\exp(y\ln x)$ whenever this is defined. This happens to coincide with the above where both are defined, namely when $x,y$ are finite positive integers. However, there are points where only one of the two definitions work, e.g., for $0^0$. This doesn't make the expression  retroactively undefined. Just like noting that often $xy=\exp(\ln x+\ln y)$ holds, doesn't make the product undefined for factors where the log is undefined.
Note however, that knowing $\lim x_n= 0$ and $\lim y_n=0$ does not allow us to conclude $\lim x_n^{y_n}=0^0=1$ or anything at all. That's why we say that $0^0$ is an indeterminate form. This is a wholly different concept.
A: $0^0$ is equal to $1$. There's too much needless confusion surrounding this. This is not a debatable point. It's not even a "convention"; it's just a fact. There is exactly one function from the empty set to the empty set. There are no ifs, ands or buts about this.
A part of the reason this "controversy" exists is because many highly voted past answers on this site insist that the value of $0^0$ is somehow up for debate or "undefined". Of course, these same people have no qualms in writing $e^x = \sum_{k \geq 0} \frac{x^k}{k!}$ for all $x \in \mathbb{R}$.
A: The expression $0^0$ is not defined.
As you state, it is clear from the definition of $a^k$ for integer $k \ge 1$:
$\begin{align*}
   a^k
     &= a                 \qquad k = 1 \\
     &= a^{k - 1} \cdot a \quad k > 1
\end{align*}$
To make $a^m \cdot a^n = a^{m + n}$ also if $m$ or $n$ are zero, we define $a^0 = 1$ as long as $a \ne 0$. To make $(a^m)^n = a^{m n}$ work out for $m$ rational, it is convenient to define $a^{1/m} = \sqrt[m]{a}$, again as long as $a \ne 0$.
To make $a^x$ continuous in $x$, we finally define $a^x = \exp(x \ln a)$. And again, this expression gets into trouble if $a = 0$.
To summarize: $0^0$ is not defined. Any definition of that will run into trouble, sooner or later.
It is common to take $0^0 = \lim_{x \to 0} x^0 = 1$ when working with series and the value turns up as the 0-th term of the series, but it is just a notational convenience (otherwise you'd have to single out the constant terms for endless hassle and no gain).
