Prove $0! = 1$ from first principles

How can I prove from first principles that $0!$ is equal to $1$?

We need $0!$ to be defined as $1$ so that many mathematical formulae work. For example we would like $$n! = n \times (n-1)!$$ to work when $n=1,$ ie $1! = 1 \times 0!.$ Also we require that the formula for the number of ways of choosing $k$ objects from $n$ is valid for $k=n.$ ie $${n \choose k} = \frac{n!}{k!(n-k)!}$$ is valid when $k=n.$

Things need to work when we extend our definition of the factorial via the gamma function.

$$\Gamma(z) = \int\limits_0^\infty t^{z-1} e^{-t} \,\mathrm{d}t,\qquad \Re(z)>0.$$

The above gives $\Gamma(n)=(n-1)!$ and so we require $0!=1,$ since $\Gamma(1)=1.$

I'm not sure that there is anything to prove. I think it follows directly from the definition of factorial:

$$ n! := \prod_{k = 1}^n k$$

So if $n=0$ the right hand side is the empty product which is $1$ by convention.

One of the simplest ways of doing this is to observe that if you have $$ 6!= 720 $$ then divide both sides by $6$ to get $$ 5!=120 $$ then divide both sides by $5$ to get $$ 4!=24 $$ then divide both sides by $4$ to get $$ 3!=6 $$ then divide both sides by $3$ to get $$ 2!=2 $$ then divide both sides by $2$ to get $$ 1!=1 $$ then divide both sides by $1$ to get $$ \text{[fill in the blank here]} $$

$0! = 1$ is consistent with, and for reasons related to, how we define the empty product.
See, for example, this entry on empty product. This is simply the name of the phenomenon Michael Hardy alludes to:

Empty product:

The empty product of numbers is the borderline case of product, where the number of factors is zero, that is, the set of the factors, is empty. In such a "borderline" case, the empty product of numbers is equal to the multiplicative identity, which is $1$.

Some of the most common examples are the following:

• The zero$^{\text{th}}$ power of a number $a$: $a^0 = 1$,
• The factorial of $0$: $0! = 1$,
• The prime factor presentation of unity, which has no prime factors.

Just as ${n^0 = 1}$ for any $n$, and the "prime factorization" of $1$ = $1$, we define, as a matter of convention, $0! = 1$.