I think, that $c=\gamma=0.57721..$ (Euler-Macheroni), 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 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/

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/