The real Gamma function has one minimum and diverges at both ends of its domain

By continuity at $1$ and [F1], $s\mathrm{\Gamma }(s)=\mathrm{\Gamma }(s+1)\to \mathrm{\Gamma }(1)=1$ as $s\downarrow 0$. Since $s\to 0^{+}$, this also gives $\mathrm{\Gamma }(s)\to +\mathrm{\infty }$.

One has $g(1)=g(2)=0$, while strict convexity [F2] gives $g(x)<0$ for $1<x<2$. Fixing such an $x$ and applying [F6] on $[1,x]$ and on $[x,2]$ gives $\xi \in (1,x)$ with $g^{\mathrm{\prime }}(\xi )<0$ and $\eta \in (x,2)$ with $g^{\mathrm{\prime }}(\eta )>0$.

By [F2] and [F5], $g^{\mathrm{\prime }}$ is nondecreasing. If $g^{\mathrm{\prime }}(u)=g^{\mathrm{\prime }}(v)$ for some $u<v$, then $g^{\mathrm{\prime }}$ is constant on $[u,v]$, so $g$ is affine there, contradicting the strict convexity of [F2]; hence $g^{\mathrm{\prime }}$ is strictly increasing and has at most one zero. By [F3] it is continuous, so the sign change in step 1.2 gives exactly one zero $s_0\in (1,2)$. Strict increase makes $g^{\mathrm{\prime }}$ negative before $s_0$ and positive after it, so [F6] makes $g$, and hence $\mathrm{\Gamma }$, decreasing on $(0,s_0]$ and increasing on $[s_0,\mathrm{\infty })$, and $s_0$ is the unique global minimum.

By [F4], $\mathrm{\Gamma }(n+1)=n!$. For $n\ge 3$, the ratios of $(n-1)!\mathrm{/}(n+1)$ grow by a factor exceeding $2$, so this sequence tends to infinity. By the eventual increase from step 2.1, if $n\le s<n+1$ then $\mathrm{\Gamma }(s)\ge \mathrm{\Gamma }(n)=(n-1)!$ and $\mathrm{\Gamma }(s)\mathrm{/}s\ge (n-1)!\mathrm{/}(n+1)$. Thus both quantities tend to infinity.