I found this proof which shows that $1! = 0$ using Taylor Series. Here's what we do: Start with the Taylor series of $\sin(x)$. We get : $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$ Now if we integrate both the sides we get: $$-\cos(x) = \frac{x^2}{2} - \frac{x^4}{4!} + \frac{x^6}{6!} - \frac{x^8}{8!} + \dots$$ So we get that $\cos(x)$ is equal to(By multiplying both sides by $-1$) : $$\cos(x) = -\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \dots$$ Now we'll again integrate both the sides, getting : $$\sin(x) = -\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \dots$$ Now if we re-arrange it, we get : $$\sin(x) = \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \dots$$ Now, we'll do the same thing with $\cos(x)$, getting : $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots (1)$$ Now, we'll re-arrange $\cos(x)$ too, getting : $$\cos(x) = -\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots (2)$$ Now, we know that $(1)=(2)$ So, $$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = -\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$$ So, we get that $1! = 0$
But, what's wrong with this proof?
Line 2 is wrong. That's not the Taylor series for $-\cos x$.
With any functions, including Taylor series, antiderivatives are only defined up to a constant. So line 2 should read $$-\cos(x) +C = \frac{x^2}{2} - \frac{x^4}{4!} + \frac{x^6}{6!} - \frac{x^8}{8!} + \dots$$ In this case $C=1$, since $\cos(0) = 1$.
The rest of the derivation covers up the basic fact that you are missing a $1$. So it's not surprising that it leads to $1=0$.