From this, we can write factorial as successive differences.
Can I use that for proving $0! = 1$ formally
The formula is as follows
$$\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$$
$l$ is any integer (don't take as zero in the below example)
Note that $r$ can be any initial value other than zero also. I am taking it as one.
Now, if I substitute $n=0$, then
$$0! = \sum_{r=0}^{0}\binom{n}{r}(-1)^r(l-r)^n = \binom{0}{0}(-1)^0 (l-0)^0 = 1(1)(1) = 1$$
There is nothing you can prove about $0!$ before you have defined it. The formula that you mentioned assumes that $n!$ is defined, for every non-negative $n$.
Here is a formal proof that uses the gamma function.
The gamma function is defined as $$\Gamma(n)=(n-1)!\tag{1}$$ where $n$ is a positive integer and $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\,dx\tag{2}$$ where $z\in\mathbb{C}$ with a positive real part.
Now consider $z=1$. From $(2)$, we have $$\Gamma(1)=\int_0^\infty x^0e^{-x}\,dx=[-e^{-x}]_0^\infty=1$$ and substituting this into $(1)$ we get that $$0!=1$$