The reflection identity of the gamma function is:
\begin{equation} \Gamma( z ) \Gamma( 1-z ) = \frac{\pi}{\sin( \pi z ) } \end{equation}
We can start from the definition of $\beta( 1- \alpha, \alpha ) $
\begin{align} \Gamma( 1- \alpha ) \Gamma( \alpha ) &= \beta ( 1- \alpha, \alpha ) = \int^\infty_0 \frac{t ^{1- \alpha-1}dt}{( 1+ t )^{1- \alpha+ \alpha} } \\ &= \int^\infty_0 \frac{t ^{-\alpha} dt}{1 + t} \\ &= \int^\infty_{-\infty} \frac{e ^{-\alpha x} e^x dx}{1+ e^x} = \int^\infty_{-\infty} \frac{e ^{( 1-\alpha )x }}{ 1 + e^{x}} = I \end{align}
where we used the substitution $t=e^x$, so $dt = e^xdx$ with limits $t=0 \implies x=-\infty$ and $t=\infty\implies x=\infty $.
Now let us integrate over the rectangular contour $C_R$ with vertices $-R, R, R + 2\pi i , - R + 2\pi i$:
\begin{align} \oint_{C_R} \frac{e ^{( 1-\alpha ) z} dz}{1 + e^z} &= \int^R_{-R} \frac{\exp( ( 1- \alpha ) x ) dx}{1 + e^x} + \int ^{2\pi}_0 \frac{\exp( ( 1-\alpha ) ( R+iy ) ) dx}{1 + \exp( R+i y ) } +\\ &+ \int^{-R}_{R} \frac{\exp( ( 1- \alpha ) ( x+2\pi i ) ) dx}{1 + \exp( x + 2\pi i ) } + \int ^{0}_{2\pi i} \frac{\exp( ( 1-\alpha ) ( -R+iy ) ) dy}{1 + \exp( -R+iy ) } \end{align}
Let us evaluate the limit $R\to\infty$ of the integrands along the sides parallel to the imaginary axis. For the first one we notice that the limit is an indeterminate form $\infty/\infty$, so we can apply L’Hôpital’s rule
\begin{align} \lim_{R \to \infty} \frac{\exp( ( 1- \alpha ) R ) \exp( ( 1- \alpha ) iy ) }{1 + \exp( R ) \exp( iy ) } &=\lim_{R \to \infty} \frac{( 1-\alpha ) \exp( ( 1-\alpha ) R ) \exp( ( 1-\alpha ) iy ) }{\exp( R ) \exp( iy ) } \\ &=\lim_{R \to \infty} ( 1-\alpha ) \exp( -\alpha R ) \exp(-\alpha iy ) = 0 \end{align}
for the second integrand it is only a matter of simplification
\begin{align} \lim_{R \to \infty} \frac{\exp( ( \alpha -1 ) R ) \exp( ( 1- \alpha ) iy ) }{1 + \exp( -R ) \exp( iy ) } &= \frac{\lim_{R \to \infty} \exp( ( \alpha -1) R ) \exp( ( 1-\alpha ) iy ) }{\lim_{R \to \infty} 1 +\exp( -R ) \exp( iy ) } \\ &=\lim_{R \to \infty} \exp( \alpha R ) \exp(-\alpha iy ) = 0 \end{align}
which also vanishes since we assume $\alpha-1<0$. So when we take the limit in the contour integral we get
\begin{align} \lim_{R \to \infty} \oint \frac{\exp( ( 1- \alpha ) z ) }{1 + \exp( z ) } &= \lim_{R \to \infty} \left\lbrace \int^R_{-R} \frac{\exp( ( 1- \alpha ) x ) dx}{1 + e^x} + \int^{-R} _{R} \frac{\exp( ( 1- \alpha ) ( x+2\pi i ) ) dx}{1 + \exp( x + 2\pi i) } \right\rbrace \\ &= \int^\infty_{-\infty} \frac{\exp( ( 1- \alpha ) x ) dx}{1 + e^x} - \int^\infty_{\infty} \frac{\exp( ( 1- \alpha ) x) \exp(( 1- \alpha ) 2\pi i ) dx}{1 + \exp( x ) \exp( 2\pi i ) } \\ &= ( 1- \exp( ( 1- \alpha ) 2\pi i ) ) \int ^\infty_{-\infty} \frac{\exp( ( 1- \alpha )x ) }{1 + \exp( x ) } \\ &= [ 1 - \exp( ( 1- \alpha )2\pi i ) ] I \end{align}
with this equality we can evaluate the contour using the residue theorem
\begin{align} \lim_{R \to \infty} \oint \frac{\exp( ( 1- \alpha ) z ) }{1 + \exp( z ) } &= 2\pi i\sum_k \text{Res}\left( \frac{\exp( ( 1- \alpha ) z ) }{1 + \exp( z ) } , z_k \right) \end{align}
It is easy to see there is only one simple pole $z_0$ inside the contour, given by $1 + \exp( z )= 0 \implies z_0= i\pi$. We evaluate its residue with L’Hôpital’s rule
\begin{align} \text{Res}\left( \frac{\exp( ( 1- \alpha ) z ) }{1 + \exp( z ) } , i\pi \right) &= \lim_{z \to i \pi} ( z-i\pi ) \frac{\exp( ( 1- \alpha ) z ) }{1 + \exp( z ) } \\ &= \lim_{z \to i\pi} \frac{\exp( ( 1-\alpha )z ) + ( z- i\pi ) ( 1- \alpha ) \exp( ( 1- \alpha ) z ) }{\exp( z ) } \\ &= \frac{e ^{i\pi} e ^{-\alpha \pi i}}{e^{i\pi}} = e ^{- \alpha \pi i}. \end{align}
We can now solve for the integral of interest $I$
\begin{align} I &= \frac{ 2\pi i e ^{- \alpha \pi i}}{1- e ^{2\pi i} e^{- \alpha \pi i}} = \pi \frac{2i }{e ^{ \alpha \pi i} - e ^{- \alpha \pi i}} \\ &= \frac{\pi}{\sin( \pi \alpha ) } \\ \therefore \Gamma( \alpha ) \Gamma( 1 - \alpha ) &= \frac{\pi}{\sin( \pi z ) } . \end{align}