What is $0!$ in base $4$?

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In base $10$ we have $0!=1$; however, I am not sure if this applies to different bases. In general, what is the 'rule' for factorials in different bases?

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Having said that $0_{10}!=1_{10}$, you can just convert the $0$ and $1$ to base $4$. What do you get? $13_4$ and $7_{10}$ are different names for the same number.

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Once $(1)_{10}=(1)_4$ then $(0!)_4=1$

The definition of factorial is the same for all bases. The only difference is the representation of the numbers.

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What is the 'rule' for factorials in different bases?

There isn't one. The factorial is just repeated multiplication, so perhaps the best way to proceed in general is to multiply in base 10 and then convert the result. Since $0! = 1$ in base 10, and $1_{10} = 1_4$, then $0!$ is still 1 in base 4. For a slightly less trivial example, let's consider $3!$ in base 4. It's $3 \cdot 2 = 6 = 1 \cdot 4^1 + 2 \cdot 4^0$, so $3! = 12_4$.

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In factorials we dont have any base system like we have in number systems.

So $(0!)_4 = 1$

In combinatorics we have factorial number system. Maybe you looking for this.

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Changing a base does not change the value of an expression. The number $16_7 = 13_{10}$ are both the same number even though they are written differently. $16_7$ is still a prime number and pile of thirteen apples isn't going to magically turn into something else just because you lost three fingers.

$16_7! = 16_7*15_7*14_7*13_7*12_7*11_7*10_7*6*5*4*3*2*1$

and $13! = 13*12*11*10*9*8*7*6*5*4*3*2*1$

and $XIII! = XIII*XII*XI*X*IX**VIII*VII*VI*V*IV*III*II*I$

are all the same thing.

So $0! = 1$. Period. So $0_b! = 1_b$ no matter what base.

... or less trivial:

$7_{10} = 10_7 = 11_6 = 12_5 = 13_4$

And $7_{10}! = 7*6*.... *2*1 = 5040_{10}$

$10_7 = 10_7*6*5*4*3*2*1 = 20460_7$

$11_6 = 11_6*10_6*5*4*3*2*1 = 35200_6$

$12_5 = 12_5*11_5*10_5*4*3*2*1 = 130130_5$

$13_4 = 13_4*12_4*11_4*10_4*3*2*1 = 1032300_4$

Which are all the exact same number, namely then number $\prod_{k=1}^{7_{10}} k$.