Let $G(z)$ be the Barnes G-function.
I want to use the infinite product representation
$$ G(z+1)=(2\pi)^{z/2}\text{exp}\left(-\frac{z(z+1)}{2}- \frac{\gamma z^{2}}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^{k}\text{exp}\left(\frac{z^2}{2k}-z\right)$$
to show that the Barnes G-function satisfies the functional equation
$$G(z+1) = \Gamma(z) G(z). $$
Specifically, I want to show that $$\frac{G(z+1)}{G(z)} = \frac{e^{-\gamma z}}{z} \prod_{k=1}^{\infty} \frac{e^{\frac{z}{k}}}{1+\frac{z}{k}}$$ where the right side of the equation is the Weierstrass infinite product representation of the gamma function.
Obviously,
$$ \begin{align} \frac{G(z+1)}{G(z)} &= \frac{(2\pi)^{z/2}\text{exp}\left(-\frac{z(z+1)}{2}- \frac{\gamma z^{2}}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^{k}\text{exp}\left(\frac{z^2}{2k}-z\right)}{(2\pi)^{(z-1)/2}\text{exp}\left(-\frac{(z-1)z}{2}- \frac{\gamma (z-1)^{2}}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z-1}{k}\right)^{k}\text{exp}\left(\frac{(z-1)^2}{2k}-(z-1)\right)} \\ &= \sqrt{2 \pi} \exp \left(-z - \gamma z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( \frac{2z-1-2k}{2k} \right). \end{align}$$
But it's not clear to me what to do next.
I will start from your last expression. The idea is, roughly, to write $(k+z-1)^k$ in the denominator as $(k+z-1)^{k-1}\cdot (k+z-1)$, then to replace $k-1$ by $k'$ and observe that the product telescopically simplifies. Also, to avoid mistakes, it is better to start with a finite product.
We have \begin{align} \prod_{k=1}^N\left(\frac{k+z}{k-1+z}\right)^k \exp\frac{2z-1-2k}{2k}=\frac{\prod_{k=1}^N(k+z)^k\exp\frac{2z-1-2k}{2k}}{\prod_{k=1}^N(k-1+z)^{k-1}(k-1+z)}=\\ =\frac{\prod_{k=1}^{N}(k+z)^k\exp\frac{2z-1-2k}{2k}}{\prod_{k=0}^{N-1}(k+z)^{k}\prod_{k=0}^{N-1}(k+z)}=\\ =\frac{(N+z)^N\prod_{k=1}^{N}\exp\frac{2z-1-2k}{2k}}{z\prod_{k=1}^{N-1}(k+z)}=\\ =\frac{N^N\left(1+\frac{z}{N}\right)^N\prod_{k=1}^{N}\exp\frac{2z-1-2k}{2k}}{\frac{z}{N+z}\prod_{k=1}^{N}k(1+\frac{z}{k})}=\\ =\frac{(N/e)^N}{N!}\times\left(1+\frac{z}{N}\right)^N\times\frac{N+z}{z}\times\prod_{k=1}^{N}\frac{e^{z/k}}{1+z/k}\times \exp\left\{-\frac12\sum_{k=1}^N\frac1k\right\}.\tag{1} \end{align} Now consider the asymptotics of different factors in the last expression as $N\rightarrow\infty$:
From Stirling's formula we get $$\frac{(N/e)^N}{N!}=\frac{1}{\sqrt{2\pi N}}+O\left(N^{-3/2}\right).$$
$\left(1+\frac{z}{N}\right)^N$ tends to $e^z$.
Using product representation of the gamma function, in the limit we can replace $\prod_{k=1}^{N}\frac{e^{z/k}}{1+z/k}$ by $ze^{\gamma z}\Gamma(z)$.
Harmonic series is known to behave as $$\sum_{k=1}^N\frac1k=\ln N+\gamma+O\left(N^{-1}\right)$$
Therefore, the limit of (1) gives $$\frac{1}{\sqrt{2\pi N}}\times e^z\times N\,e^{\gamma z}\Gamma(z)\times \frac{e^{-\gamma/2}}{\sqrt{N}}\sim\frac{\exp\left\{z(\gamma+1)-\frac{\gamma}{2}\right\}}{\sqrt{2\pi}}\Gamma(z).$$ This is equivalent to the functional relation you want to prove.