Zero to the Power of Zero

Examples
In light of some mathematicians' reticence to define the zeroth power of zero as $1$, the following are examples of reasons why defining $0^0 = 1$ is a good idea.

Empty Product
We can interpret $0^0$ as meaning "zero multiplied by itself zero times".

Using product notation:


 * $0^0 = \displaystyle \prod_{\large \text{false}} 0$

This is a vacuous product, so by definition should be equal to $1$.

Binomial Theorem
Consider the real polynomial function:


 * $y = \left({x + c}\right)^n$

for $n \in \N, c \in \R$.

By the binomial theorem, $y$ contains a term of the form:


 * $\dbinom n n x^{n - n} c^n$

If we did not define $0^0 = 1$, $y$ would have a discontinuity at $x = 0$.

This would contradict the theorem that a polynomial is continuous on the entire real line.

Cardinality of Mappings
By Cardinality of Set of All Mappings, the number of mappings from the empty set to the empty set should be given by:

By Empty Mapping is Unique, there is exactly $1$ such mappings, demanding that $0^0 = 1$.

Exponential of Zero
By Exponential of Zero:


 * $\exp 0 = 1$

From Power Series Expansion for Exponential Function


 * $\exp x = \dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \cdots$

For these theorems to be consistent, it is necessary that:


 * $\exp 0 = 1 = \dfrac {0^0} {0!} + 0 + 0 + \cdots$

which holds only if $0^0 = 1$.

Derivatives
Consider the identity mapping:


 * $I_{\mathbb F} \left({x}\right) = x$

where $\mathbb F \in \left\{ {\R, \C}\right \}$.

From Derivative of Identity Function:


 * $\dfrac {\mathrm d I_{\mathbb F} }{\mathrm d x} = 1$

But $I_{\mathbb F} \left({x} \right) = x^1$ is also an order one polynomial.

By Power Rule for Derivatives:


 * $\dfrac {\mathrm d I_{\mathbb F} } {\mathrm d x} = 1 x^0$

As $I_{\mathbb F}$ is differentiable at $0$, for these theorems to be consistent, we insist that $0^0 = 1$.

As a Limit
Consider the real function:


 * $y = x^x$

This function is well defined for $x > 0$.

It is not obvious whether or not the right hand limit:


 * $\displaystyle \lim_{x \mathop \to 0^+} y$

exists.

If it does, it would be nice if:


 * $\displaystyle \lim_{x \mathop \to 0^+} x^x = 0^0$

Indeed, by Limit of x to the x, we have:


 * $\displaystyle \lim_{x \mathop \to 0^+} x^x = 1$

We see that defining $0^0 = 1$ allows $x^x$ to be right-continuous at $x = 0$.