The rule for finding the remainder when dividing by $3$ is to sum up the digits and divide THAT number by $3$, the remainders will be the same. As $14!$ is divisible by $3$ the remainder should be zero, so
$$8 + 7 + 1 + a + 8 + 2 + b + 1 + 2 = 29 + a + b$$
should be divisible by $3$. To make it easier we can take factors of $3$ out of the $29$ and conclude that $2 + a + b$ should be divisible by $3$.
The rule for remainders when dividing by $9$ is the same, sum up the digits and divide THAT by $9$. There's also a rule for $11$ involving the alternating sum of digits. All these give you equations like the one I got above. Try and write down those equations and see if you can do the last step of solving them on your own.
To answer your question, there are several methods that can help calculate factorials faster, though most are either approximations or shortcuts for calculation. I will be detailing some of these methods below:
Method 1
As mentioned by Joe in the comments, Stirling's approximation is a good method to approximate the value of a large factorial, and by rewriting the factorial as a Gamma function, the following formula is obtained:
$$n! \sim \sqrt{2\pi n} \cdot \biggl(\frac{n}{e} \biggr)^n$$
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Method 2
Again, as mentioned by Math101 in the comments, we can make use of the property $n! = n(n - 1)!$ to quickly calculate a factorial given the value of a previous factorial.
For example, given that $5! = 120$ (or by calculating $5!$ by using the same method):
$$6! = 6 \cdot 5! = 6 \cdot 120 = 720$$
This method does not work as efficiently as Method 1, especially at larger values of $n$, but it may be useful in some ways.
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Method 3
By pairing up the numbers needed to calculate a factorial, a shortcut can be used to quickly calculate the factorial by using the commutative property ($xy = yx$).
For example:
$$8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = (8 \cdot 1) \cdot (7 \cdot 2) \cdot (6 \cdot 3) \cdot (5 \cdot 4) = 8 \cdot 14 \cdot 18 \cdot 20 = 40320$$
The largest value is always paired up with the smallest value, the second-largest with the second-smallest, and so on.
Please have a read through the source for this method for more detail on how to use this method more efficiently.
Please note that there is no clear method to easily calculate large factorials; approximations can only bring you fairly close to the true value.
I hope this helps! I'm still new to MSE, so I would really appreciate any and all feedback. Thank you!
Best Answer
Converting/collating comments:
You may of course define anything you like. Whether it should have the same name as something else common in the literature and whether or not it is useful is an entirely different story.
If we were to want to extend the factorial function, there are a few properties we would surely want. First, that the values agree for each natural number. Your extension does accomplish this. Second, that we would have $x! = x\times (x-1)!$. Again, your extension does accomplish this, so good job on that (assuming you take care to be referring to the decimal expansion of $x-\lfloor x\rfloor$ to avoid issues when moving to negative numbers).
Additional properties that we would likely want would be for it to be increasing over the reals $[1,\infty)$ and continuous. Your extension fails to be here noting that $(1.99)! = 1!\times 1.9\times 1.09 = 2.071>2!$ and further that $\lim\limits_{x\to 2} x!$ doesn't exist since from the right it would be $2.00000\dots$ which would equal $2$ but from the left it would be $1.99999\dots$ which would be $\prod\limits_{k=1}^\infty (1+9\times 10^{-k})\approx 2.0917288\dots$
The most common extension to the factorial function would be the Gamma Function (with an offset). This function has the desired properties plus many more. It extends the factorial to the entire complex plane (except non-positive integers which are poles), it is analytic, meromorphic, holomorphic, and has other nice properties. Of course, many extensions can exist, but this is the extension which has been the most useful for applications.