We could use the Log-Gamma series representation. For $x> 0$ : $$ \ln{\Gamma\left(x\right)}=-\ln{x}-\gamma x+\sum_{n=1}^{+\infty}{\left(\frac{x}{n}-\ln{\left(1+\frac{x}{n}\right)}\right)} $$
for which an elementary proof could be done by considering the integral $ I_{n}\left(x\right)=\int_{0}^{1}{\left(1-t\right)^{n}t^{x-1}\,\mathrm{d}t} $.
Define $ u_{n}:x\mapsto\frac{x}{n}-\ln{\left(1+\frac{x}{n}\right)} $. $ \sum\limits_{n=1}^{+\infty}{u_{n}} $ is well defined on $ \left(0,+\infty\right) $, because $ u_{n}\left(x\right)=\underset{n\to+\infty}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) $. $$ $$
Substituting $ t=\frac{u}{n} $, gives $ I_{n}\left(x\right)=n^{-x}\int_{0}^{n}{\left(1-\frac{u}{n}\right)^{n}u^{x-1}\,\mathrm{d}u}=n^{-x}\int_{0}^{+\infty}{g_{n}\left(u\right)u^{x-1}\,\mathrm{d}u} $, where $ g_{n} $ is the function defined as $ g_{n}:x\mapsto\left\lbrace\begin{matrix}\left(1-\frac{x}{n}\right)^{n} & 0\leq x\leq n\\ 0 & x\geq n\end{matrix}\right. $.
$g_{n}$ can be proven to converge uniformly to $ \exp $, which leads to the sequence of functions $ x\mapsto n^{x}I_{n} $ converging to $ \Gamma $, and thus $ x\mapsto\ln{\left(n^{x}I_{n}\left(x\right)\right)} $ converging to $ \ln\Gamma $. $$ $$
Now given $ n\geq 1 $, $ x > 0 $, and $ k\geq 1 $, \begin{aligned}\small I_{n-k}\left(x+k\right)=\int_{0}^{1}{\left(1-t\right)^{n-k}t^{x+k-1},\mathrm{d}t}&\small =\left[\frac{\left(1-t\right)^{n-k}t^{x+k}}{x+k}\right]{0}^{1}+\frac{n-k}{x+k}\int{0}^{1}{\left(1-t\right)^{n-k-1}t^{x+k},\mathrm{d}t}\ \small I_{n-k}\left(x+k\right)&\small =\frac{n-k}{x+k}I_{n-k-1}\left(x+k+1\right)\ \small\Longrightarrow\prod_{k=0}^{n-1}{\frac{I_{n-k}\left(x+k\right)}{I_{n-k-1}\left(x+k+1\right)}}&\small =\prod_{k=0}^{n-1}{\frac{n-k}{x+k}}\ \small\iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I_{n}\left(x\right)&\small =\frac{n!}{\prod\limits_{k=0}^{n}{\left(x+k\right)}}=\frac{n^{-x}}{x}\left(1+\frac{1}{n}\right)^{-x}\prod_{k=1}^{n}{\frac{\left(1+\frac{1}{k}\right)^{x}}{1+\frac{x}{k}}}\end{aligned}
Hence : $$ \small\ln{\left(n^{x}I_{n}\left(x\right)\right)}=-\ln{x}-x\ln{\left(1+\frac{1}{n}\right)}-x\sum_{k=1}^{n}{\left(\frac{1}{k}-\ln{\left(1+\frac{1}{k}\right)}\right)}+\sum_{k=1}^{n}{\left(\frac{x}{k}-\ln{\left(1+\frac{x}{k}\right)}\right)} $$
Taking the limit and identifying it with the previously established one, gives the desired result.
Integrating that formula : \begin{aligned} \small\int_{0}^{1}{\ln{\Gamma\left(x\right)}\,\mathrm{d}x}&\small =1-\frac{\gamma}{2}+\sum_{n=1}^{+\infty}{\int_{0}^{1}{\left(\frac{x}{n}-\ln{\left(1+\frac{x}{n}\right)}\right)\mathrm{d}x}}\\ &\small =1-\frac{\gamma}{2}+\sum_{n=1}^{+\infty}{\left(\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}-\frac{1}{2n}-n\ln{\left(1+\frac{1}{n}\right)}+1\right)}\\ &\small =1-\frac{\gamma}{2}+\color{blue}{\sum_{n=1}^{+\infty}{\left(\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}\right)}}-\color{red}{\sum_{n=1}^{+\infty}{\left(\frac{1}{2n}+n\ln{\left(1+\frac{1}{n}\right)}-1\right)}} \end{aligned}
The blue sum evaluates to $ \gamma $ using our first formula. For the red sum, we want to use the power series for $ \ln $ and switch the order of the summations using fubini theorem for summable families, and that would give $ \sum\limits_{n=2}^{+\infty}{\frac{\left(-1\right)^{n}\zeta{\left(n\right)}}{n+1}} $.
The integral can be evaluated the following way : \begin{aligned}\small 2\int_{0}^{1}{\ln{\Gamma\left(x\right)}\,\mathrm{d}x}&\small =\int_{0}^{1}{\ln{\Gamma\left(x\right)}\,\mathrm{d}x}+\int_{0}^{1}{\ln{\Gamma\left(1-x\right)}\,\mathrm{d}x}\\ &\small =\int_{0}^{1}{\ln{\left(\frac{\pi}{\sin{\left(\pi x\right)}}\right)}\,\mathrm{d}x}\\ \small 2\int_{0}^{1}{\ln{\Gamma\left(x\right)}\,\mathrm{d}x}&\small =\ln{\pi}-\frac{2}{\pi}\color{blue}{\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{x}\right)}\,\mathrm{d}x}}\end{aligned}
The blue integral is a classic and can be evaluated to $ -\frac{\pi}{2}\ln{2} $ either using Riemann sums or a couple of tricky substitutions. Thus $ 2\int_{0}^{1}{\ln{\Gamma{\left(x\right)}}\,\mathrm{d}x}=\ln{\left(2\pi\right)} $.
Hence the result.
