I think, that $c=\gamma=0.57721..$ (Euler-Mascheroni), because
$$
\sum_{k=1}^{n-1} \frac{1}{(n-k) c + \ln \Gamma(n+1)-\ln \Gamma(k+1)} = \\
\sum_{m=1}^{n-1} \frac{1}{\ln(\frac{ e^{m c}\  \Gamma(n+1)}{\Gamma(n-m+1)})} \sim  
\sum_{m=1}^{n-1} \frac{1}{m \ \ln (e^{c}\  n)} = \\
\frac{1}{\ln (e^{c}\  n)} H_{n-1} \sim \frac{1}{\ln (e^{c}\  n)} (\ln n +\gamma)
$$
where in the second line I changed the summation variable to $m=n-k$ and used
$$
\frac{\Gamma(x+a)}{\Gamma(x+b)}\sim x^{a-b}
$$
for large $x$.$H_n$ is the $n$-th Harmonic number and its asymptotic expansion for large $n$ is
$$
H_{n}\sim \ln n +\gamma +O(n^{-1}).
$$
All expansions/definitions can be found at Wikipedia or at https://dlmf.nist.gov/