Use the fact that $\Gamma (1-z) = -z\, \Gamma(-z)$ and then:
$$\Gamma(1-z)\Gamma(z) = -z \, \Gamma(-z)\Gamma(z) = -z \cdot \frac{1}{-z}\cdot \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{\left(1 + \frac{z}{k} \right)\left(1 - \frac{z}{k} \right) } = \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{1 - \frac{z^2}{k^2}} = \frac{\pi}{\sin \pi z}$$
Unfortunately, this is not new, though I would like to offer another derivation;
\begin{align*}
\int_{0}^{t}H_ydy &= \int_{0}^{t}\int_{0}^{1}\frac{1-x^y}{1-x}dxdy\\
&= \int_{0}^{1}\int_{0}^{t}\frac{1-x^y}{1-x}dydx\\
&= \int_{0}^{1}\frac{t}{1-x}+\frac{1-x^t}{(1-x)\ln(x)}dx\\
(1)&= \int_{0}^{1}\frac{t}{1-x}+\sum_{j=0}^{t-1}\frac{x^j}{\ln(x)}dx\\
&=\lim_{x\rightarrow 1^{-}}\left( -t\ln(1-x)+\sum_{j=0}^{t-1}\text{li}(x^{j+1})\right)\\
(2)&=\lim_{x\rightarrow 1^{-}}\sum_{j=0}^{t-1}\left(\text{li}(x^{j+1})-\ln(1-x)\right)\\
&= \gamma t + \sum_{j=0}^{t-1}\ln(j+1)\\
&= \gamma t + \ln(t!)\\
&= \gamma t +\ln\Gamma(t+1)
\end{align*}
Now writing $H_y$ as $\sum_{n=1}^{\infty}\frac{y}{n(n+y)}$ gives
\begin{align*}
\int_{0}^{t}H_ydy &= \gamma t + \ln\Gamma(t+1)\\
&= \int_{0}^{t}\sum_{n=1}^{\infty}\frac{y}{n(n+y)}dy\\
&= \sum_{n=1}^{\infty}\frac{t}{n}-\ln \left(1+\frac{t}{n}\right)
\end{align*}
solving for $\Gamma(t+1)$ and using $\Gamma(t+1) = t\Gamma(t)$ gives
$$\Gamma(t) = \frac{e^{-\gamma t}}{t}\prod_{n=1}^{\infty}\left(1+\frac{t}{n}\right)^{-1}e^{\frac{t}{n}}$$
EDIT
First, I should say that I am just starting to study analysis - I just try stuff and see if it works, so I apologize for the lack of rigour.
I'm not too sure which steps I should fill in, so I will try to fill in any apparent holes.
assuming we only know that $\gamma$ shows up as
$$\lim_{N \rightarrow \infty}(H_N - \ln(N)) = \gamma,$$
one can show that $\int_{0}^{1}H_ydy=\gamma$ using $H_y=\sum_{k=2}^{\infty}(-1)^k\zeta(k)y^{k-1}$ for $|y|<1$ (I have a crude derivation here).
Knowing that, we get
\begin{align*}
\gamma = \int_{0}^{1}H_ydy &= \int_{0}^{1}\int_{0}^{1}\frac{1-x^y}{1-x}dxdy\\
&= \int_{0}^{1}\int_{0}^{1} \frac{1}{1-x}-\frac{x^y}{1-x}dydx\\
&= \int_{0}^{1}\frac{1}{1-x}+\frac{1}{\ln(x)}dx\\
(*)&= \lim_{x\rightarrow 1^{-}}(\text{li}(x)-\ln(1-x)),\\
\end{align*}
where $\text{li}(x) = \int_{0}^{x}\frac{1}{\ln(t)}dt$ is the logarithmic integral.
Now we need to evaluate $\int_{0}^{t}\frac{x^k}{\ln(x)}dx$ for (1); Make the substitution $\;x=e^u$ to get
$$\int_{0}^{t}\frac{x^k}{\ln(x)}dx = \int_{-\infty}^{\ln(t)}\frac{e^{(k+1)u}}{u}du,$$
now let $v=(k+1)u$ and see that
\begin{align*}
\int_{0}^{t}\frac{x^k}{\ln(x)}dx &= \int_{-\infty}^{(k+1)\ln(t)}\frac{e^x}{x}dx\\
&= \int_{0}^{t^{k+1}}\frac{1}{\ln(x)}dx\\
&=\text{li}(t^{k+1})
\end{align*}
after the substitution $x=\ln(u)$.
Next we need to evaluate $\lim_{x\rightarrow 1^{-}}(\text{li}(x^k)-\ln(1-x))$.
so by using $(*)$ -
\begin{align*}
\lim_{x\rightarrow 1^{-}}(\text{li}(x^k)-\ln(1-x)) &= \lim_{x\rightarrow 1^{-}}(\text{li}(x) - \ln(1-\sqrt[k]{x})\\
&= \lim_{x\rightarrow 1^{-}}\left(\text{li}(x) - \ln \left(\frac{1-x}{1+\sqrt[k]{x} + \cdots +\sqrt[k]{x^{k-1}}}\right)\right)\\
&= \gamma + \ln(k)
\end{align*}
I just have to say that I really like how you derived it.
Best Answer
Euler, of course, was prolific in his investigation of infinite series and products.
Supposedly, he examined an infinite product with a salient pattern
$$\left(\frac{2^n}{1^n}\frac{1}{n+1}\right)\left(\frac{3^n}{2^n}\frac{2}{n+2}\right)\left(\frac{4^n}{3^n}\frac{3}{n+3}\right)\ldots \ (*)$$
and observed that if $n$ is a positive integer, then the product converges to $n!$.
Using your notation ($s \in \mathbb{N})$, the partial product for $(*)$ can be expressed in the form where $(n+1)^s$ appears
$$P_n(s)=\frac{n!}{(s+1)(s+2)\ldots(s+n)}(n+1)^{s}.$$
After some cancelation and manipulation we get for $n >s,$
$$P_n(s)=s!\left(\frac{1+1/n}{1+2/n}\right)\left(\frac{1+1/n}{1+3/n}\right)\ldots\left(\frac{1+1/n}{1+s/n}\right),$$
and it is apparent that $P_n(s) \rightarrow s!$ as $n\rightarrow \infty.$