How to prove that $ \ln{G(z+n)}+\ln{Γ(z+n)}=\ln{G(z)} +(n+1)\ln{Γ(z)}+\displaystyle \sum_{k=1}^{n} k\ln(z+n-k) $? where Γ(z) is the gamma function, G(z) is Barnes G Function. In Barnes' original 1857 article on the G function, on page 266 of this issue of "The Quarterly Journal Of Pure And Applied Mathematics", he starts off by stating an equality equivalent to the one in the title. I'm trying to understand how he arrived at his definition of the G function and this already got me baffled.

Recall that $G(z+1)=\Gamma(z)G(z)$ and $\Gamma(z+1)=z\Gamma(z)$. Repeatedly applying the functional equation for the G-function yields
$$G(z+n)\Gamma(z+n) =\Gamma(z+n)\Gamma(z+n-1)\Gamma(z+n-2)\cdots\Gamma(z)G(z).\qquad(1)$$
Similarly, repeatedly applying the functional equation for the gamma function gives
$$\Gamma(z+l)=\Gamma(z)\prod_{j=0}^{l-1}(z+j).$$
We may replace each gamma function (except for the rightmost $\Gamma(z)$) in the RHS of $(1)$ with the above product to acquire
\begin{align}
G(z+n)\Gamma(z+n) &=G(z)\Gamma(z)\prod_{l=1}^n\left(\Gamma(z)\prod_{j=0}^{l-1}(z+j)\right), \\
&\qquad\text{expand out the double product and} \\
&\qquad\text{ collect similar factors} \\
&= G(z)\Gamma^{n+1}(z)\prod_{k=1}^n(z+n-k)^k.
\end{align}
So we have
$$G(z+n)\Gamma(z+n)=G(z)\Gamma^{n+1}(z)\prod_{k=1}^n(z+n-k)^k$$
and taking the logarithm gives you what you're after.
