First, we know that (except for n=0), there are more factors of 2 than factors of 5 in $n!$, so that the first non zero digit has to be even, and we only need to know what it is modulo $5$.
Define $\phi : \mathbb{N}^* \to (\mathbb{Z}/5\mathbb{Z})^*$ by $\phi(n) = {\bar3}^{v_5(n)} \times d(n)$ where $v_5(n)$ is the number of $5$ in the factorisation of $n$ and $d(n)$ is the class of the last nonzero digit of $n$ when written in base $5$. The goal is to find $\phi(n!)$.
It turns out that $\phi$ is a group morphism from $(\mathbb{N}^*,\times)$ to $((\mathbb{Z}/5\mathbb{Z})^*,\times)$ :
If $n = 5^k(a+5b)$ and $m = 5^{k'}(a'+5b')$ with $a$ and $a'$ coprime with $5$, then $nm = 5^{k+k'}(aa'+5(ab'+ba'+5bb'))$, thus $\phi(nm) = 3^{k+k'}aa' = (3^k a)(3^{k'}a') = \phi(n)\phi(m)$.
Therefore we only need to find $\phi(k)$ for $k=1 \ldots n$ and multiply them together to get $\phi(n!)$.
If $k$ is coprime with $5$, then $\phi(k) = \bar{k}$, and $\phi(k) = 3 \phi(k/5)$ otherwise.
Furthermore, $1*2*3*4 = 4$ in $(\mathbb{Z}/5\mathbb{Z})^*$ so if $n=5a$ :
$$\phi(n!) = \left(\prod_{i=0}^{a-1} \phi(5i+1)\ldots\phi(5i+5)\right)
= \left(\prod_{i=0}^{a-1} 3 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot \phi(i+1)\right) \\
= \left(\prod_{i=0}^{a-1} 2 \phi(i+1)\right) = 2^a\phi(a!)$$
If $n=5a+b$, then $\phi(n!) = \phi((5a)!)\phi(5a+1)\ldots\phi(5a+b) = \phi((5a)!)\phi(1)\ldots\phi(b) = \phi((5a)!)\phi(b!)$
Therefore, we have the recurrence relation $\phi(n!) = 2^{[n/5]} \phi([n/5]!) \phi((n \mod 5)!)$
Now to get the digit in base $10$ : if $n! = 10^k (a + 10 \ldots)$ with $a \in \{2;4;6;8\}$ then, in base $5$, we get $n! = 5^k ((2^k a) + 5 \ldots)$, so $\phi(n!) = 3^k*2^k * a = a$ in $(\mathbb{Z}/5\mathbb{Z})^*$
So we simply need to look at $\phi(n!)$ and pick the corresponding even digit :
In fact, $((\mathbb{Z}/5\mathbb{Z})^*= \{1;2;3;4\},\times)$ is isomorphic to $(\{6;2;8;4\},\times)$ where the multiplication is modulo $10$.
Rewriting the recurrence relation in this context, we get :
$D(n) = 2^{[n/5]} D([n/5]) D(n \mod 5)$ where the multiplications are all modulo $10$.
To recover the recurrence relation you have, we only need to prove that $2^{[n/5]} D(n \mod 5) = 4^{[n/10]} D(n \mod 10)$ :
If the last digit of n is less than $5$, then $[n/5] = 2[n/10]$ and $n \mod 5 = n \mod 10$, so they are equal.
If not, then $[n/5] = 2[n/10] +1$ and $D(n \mod 10) = D(5) D(n \mod 5) = 2 D(n \mod 5)$ so they are equal again.
$$v_p(n!)=\sum_{k\ge 1} \lfloor \frac{ n }{p^k} \rfloor = \sum_{k=1}^{\lfloor \log_p(n) \rfloor} (\frac{ n }{p^k}-O(1))=\frac{ n}{p} \frac{1-p^{-\lfloor \log_p(n) \rfloor}}{1-p^{-1}} - O(\lfloor \log_p(n) \rfloor)$$ $$ \in [\frac{ n}{p} \frac{1-p^{-\lfloor \log_p(n) \rfloor}}{1-p^{-1}} - \lfloor \log_p(n) \rfloor,\frac{ n}{p} \frac{1-p^{-\lfloor \log_p(n) \rfloor}}{1-p^{-1}}]$$
That's quite the best possible approximation because with $n= p^k$ then $v_p(n!)-v_p((n-1)!) = \lfloor \log_p(n) \rfloor$
Best Answer
In this answer, it is shown that the number of factors of $p$ in $n!$ is $$ \frac{n-\sigma_p(n)}{p-1}\tag{1} $$ where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.
Factor $$ 504=2^3\cdot3^2\cdot7\tag{2} $$ Write $10200$ in base-$2$, base-$3$, and base-$7$: $$ \begin{array}{}10011111011000_2&111222210_3&41511_7\end{array}\tag{3} $$ The number of factors of $2$ in $10200!$ is $\frac{10200-8}{2-1}=10192$.
The number of factors of $3$ in $10200!$ is $\frac{10200-12}{3-1}=5094$.
The number of factors of $7$ in $10200!$ is $\frac{10200-12}{7-1}=1698$.
Since $\left\lfloor\frac{10192}{3}\right\rfloor=3397$, $\left\lfloor\frac{5094}{2}\right\rfloor=2547$, and $\left\lfloor\frac{1698}{1}\right\rfloor=1698$, the maximum value of $n$ so that $\frac{10200!}{504^n}$ is an integer is $n=1698$.
To answer the question asked:
For large $n$, the sum of the digits of $n$ is small compared to $n$, so suppose $$ d=p_1^{e_1}p_2^{e_2}\dots p_m^{e_m}\tag{4} $$ Following the computations above, the greatest power of $d$ that divides $n!$ is $$ \min_k \left\lfloor\frac{n-\sigma_{p_k}(n)}{e_k(p_k-1)}\right\rfloor\tag{5} $$ Ignoring $\sigma_{p_k}(n)$ as negligible, the greatest of $e_k(p_k-1)$ is a strong indicator of which $p_k$ is the constraining factor.
In the current case, the greatest of $3(2-1)=3$, $2(3-1)=4$, and $1(7-1)=6$ hints strongly that $7$ is the constraining factor.