Is it a question or a murder weapon?

$First\quad we\quad take\quad P=log\left( \Gamma \left( 1+x \right) \right) \\ Using\quad taylor\quad series\quad of\quad expansion,\\ P=-\gamma x+\sum _{ k=2 }^{ \infty }{ \left( \frac { \zeta \left( k \right) }{ k } .{ \left( -x \right) }^{ k } \right) } \\ where\quad \gamma \quad is\quad Euler-Mascheroni\quad constant\\ and\quad \zeta (k)=\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ k } } } \\ Now\quad on\quad using\quad the\quad integral\quad forms\quad of\quad \gamma \quad and\quad \zeta (k),\\ we\quad get\\ P=\int _{ 0 }^{ \infty }{ \left( \frac { { e }^{ -\left( x \right) t }-\left( x \right) { e }^{ -t }-1+\left( x \right) }{ t\left( { e }^{ t }-1 \right) } \right) dt } \\ Now\quad on\quad substituting\quad x\quad =\quad \frac { r }{ 24 } -1,\quad we\quad get\\ P=\int _{ 0 }^{ \infty }{ \left( \frac { { e }^{ -\left( \frac { r-24 }{ 24 } \right) t }-\left( \frac { r-24 }{ 24 } \right) { e }^{ -t }-1+\left( \frac { r-24 }{ 24 } \right) }{ t\left( { e }^{ t }-1 \right) } \right) dt } \\ \therefore \quad \prod _{ r=1 }^{ 23 }{ \left( { e }^{ \left( \int _{ 0 }^{ \infty }{ \left( \frac { { e }^{ -\left( \frac { r-24 }{ 24 } \right) t }-\left( \frac { r-24 }{ 24 } \right) { e }^{ -t }-1+\left( \frac { r-24 }{ 24 } \right) }{ t\left( { e }^{ t }-1 \right) } \right) dt } \right) } \right) } =\prod _{ r=1 }^{ 23 }{ \left( { e }^{ \left( P \right) } \right) } \\ =\quad \prod _{ r=1 }^{ 23 }{ \left( { e }^{ \left( log\left( \Gamma \left( \frac { r }{ 24 } \right) \right) \right) } \right) } \\ =\prod _{ r=1 }^{ 23 }{ \left( \Gamma \left( \frac { r }{ 24 } \right) \right) } =\prod _{ r=1 }^{ 23 }{ \left( \Gamma \left( \frac { r }{ 23+1 } \right) \right) = } \sqrt { \frac { { \left( 2\pi \right) }^{ 23 } }{ 24 } }$

The proof of the last step can be seen from here .

Thanks to Kartik, Ishan and Parth.

This is a great problem! I loved solving it.

So we have to decode this murder weapon -

$\displaystyle \int_0^\infty \dfrac{e^{-\left(\frac{r-24}{24} \right)t} - \left(\frac{r-24}{24} \right)e^{-t} -1 + \left(\frac{r-24}{24} \right)}{t (e^t-1)} \, dt$

Let's consider the following integral

$\displaystyle I(n) = \int_0^\infty \dfrac{e^{-nt} - n e^{-t} -1 + n}{t(e^t -1)} \ dt$

$\displaystyle I'(n) = \int_0^\infty \dfrac{-t e^{-nt} - e^{-t} +1}{t(e^t -1)} \ dt$

$\displaystyle I'(n) = \int_0^\infty \left(\dfrac{e^{-t}}{t} - \dfrac{e^{-nt}}{e^t-1}\right)\ dt$

$\displaystyle I'(n) = \int_0^\infty \left(\dfrac{e^{-t}}{t} - \dfrac{e^{-(n+1)t}}{1-e^{-t}}\right)\ dt$

Now, this is probably Aditya Kumar 's favorite integral - he was really excited about it when he posted it. It simply evaluates to

$\displaystyle I'(n) = \psi(n+1)$ , the digamma function

$\displaystyle \int_0^n {I'(n) \ dn} = \int_0^n {\psi(n+1) \ dn}$

$\displaystyle I(n) = \ln(\Gamma(n+1))$

and $\displaystyle I\left(\frac{r-24}{24}\right) = \ln\left(\Gamma\left(\frac{r-24}{24} + 1\right)\right)$

$\displaystyle = \ln\left(\Gamma\left(\frac{r}{24}\right)\right)$

and

$\displaystyle \prod_{r=1}^{23}{\Gamma\left(\frac{r}{24}\right)} = \boxed{\sqrt{\frac{{(2\pi)}^{23}}{24}}}$