Simple Factorial Challenge In light of today's date...
A factorial of a number n, is the product of all the numbers from 1 to n inclusive.
The Challenge
Given an integer n where 0 &lt;= n &lt;= 420, find the sum of the digits of n!. It's that simple :P
Notes: I feel like this challenge isn't too simple, as some languages have a very hard time storing numbers that are fairly large. Also, coding a BigInteger datatype is fairly straightforward, but hard to golf.

Examples

Input: 100. Output: 648.
Input: 200. Output: 1404.
Input: 420. Output: 3672.


Scoring

The input and output can be given by any convenient method.
Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
Standard loopholes are forbidden.
This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

EDIT: Ok, I guess it's actually easy...take a look at the Jelly answer below. Anyways, I guess we all know which one's going to be accepted. But, I'd still like to see some interesting ideas and tricks to this problem!
(old problem) JavaScript (Node.js), 99 94 92 88 86 83 80 78 bytes



f=(x,t=[...21+Array(999)])=&gt;x?f(s=x-1,t.map(d=&gt;-s+(s+=(t=~~d*x+t/10|0)%10))):s


Try it online!

well, first language without large number support

Thank Arnauld for 4 bytes



f = (x, t = ['2', '1', ...','.repeat(998)]) =&gt; // init x, t=1
  x ?
    f( // recursive x':=x-1, t'=t*x
      s = x-1, // s init to 0 when x==1, in next recursion just return
      t.map (
        b = // init b to NaN
        d =&gt;
          -s + ( //  to return the added value
            s+= ( // -s + (s + k) == k
              b=
                ~~d*x+b/10 | 0
                // ~~d converts ',' to 0
                // b is initally NaN, after first iteration
                // it's zero, then the last digit sum result with carry
            ) % 10
          )
      )
    )
:
  s // the sum from x==1