Imaginary Factorial

$i! \\$ $= \Gamma\left(1+i\right)\\$ $= \int_0^{\inf} e^{-t}t^i dt \\$ $= \int_0^{\inf} e^{-t}e^{i log(t)} dt \\$ $=\int_0^{\inf} e^{-t}\left(cos\left(log\left(t\right)\right)+i sin\left(log\left(t\right)\right)\right)dt\\$ $=\int_0^{\inf} e^{-t}cos\left(log\left(t\right)\right)dt + i \int_0^{\inf} e^{-t}sin\left(log\left(t\right)\right)dt\\$
Hence it's a complex number.

We can use Gamma Function ,

$\Gamma (n)=(n-1)!$ .

In this case, $n-1=i\Rightarrow n=1+i.$

$\therefore i!=\Gamma (1+i)$ , which is a complex number, according to our calculators (currently I can only think of this...) :/

My method is quite raw,but I just used logic: Definition of n!=1x2x3x...x(n-1)x(n) So i!=1x2x3x...x(i-1)x(i), which we can rewrite as Cx....(i-1)x(i) where C is just a constant we get from multiplying all numbers 1x2x3...etc. Assume i!:=Cx(i) => holds as Ci is a complex number. Now assume i!:=Cx(i-1)x(i) => Cx(i^2-i) = Cx(-1-i) which is also complex number. We can spot the pattern that no matter how many terms we have (i)x(i-1)x(i-2)...(i-m) for all m, we will eventually get complex number of we expand the brackets. Hence simplify i!=Cx(a+bi) and knowing that complex numbers are closed under multiplication ( in this case by some real number C) we get i!=a+bi for arbitrary(but non zero) a and b.

The factorial denotes !. So it's like 1 x 2 x 3 x... When multiplied to i, it becomes unreal number still.