A Gamma limit $\lim_{n\rightarrow+\infty}\sum_{k=1}^n \left( \Gamma\bigl(\frac{k}{n}\bigr)\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}$
Step 1. If $$I_1(n)=\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$ Then $\lim\limits_{n\to\infty}I_1(n)=0$.
Proof. Indeed, since $\Gamma$ is decreasing on $(0,1]$ we have $$ I_1(n)\leq\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}\leq\sum_{k=1}^\infty\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}=\frac{1}{\Gamma(1/\sqrt{n})-1}$$ and step 1. follows.
Step 2. If $$I_2(n)=\sum_{\sqrt{n}<k\leq n/2}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$ Then $\lim\limits_{n\to\infty}I_2(n)=0$.
Proof. Recall that $\Gamma$ attains its minimum $\approx0.8856$, on $[1,2]$, at some some point $x_0\approx1.4616$. In particular, $\Gamma(x)\geq2/3$ for $1\leq x\leq 2$. So, for $\sqrt{n}<k\leq n/2$ we have $$ \frac{k}{n}\Gamma\left(\frac{k}{n}\right)=\Gamma\left(1+\frac{k}{n}\right) \geq\frac{2}{3} $$ Thus, for $\sqrt{n}<k\leq n/2$, we have $\Gamma(k/n)>4/3$. It follows that $$ I_2(n)\leq \sum_{k>\sqrt{n}}\left(\frac{3}{4}\right)^k=4\left(\frac{3}{4}\right)^{\lceil\sqrt{n}\rceil} $$ and step 2. follows.
Step 3. If $$I_3(n)=\sum_{n/2<k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$ Then $\lim\limits_{n\to\infty}I_3(n)=\dfrac{e^\gamma}{e^\gamma-1}$. where $\gamma$ is the Euler-Mascheroni constant.
Proof. Note first that, with $p=n-k$, $$ I_3(n)=\sum_{0\leq p<n/2}\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n} =\sum_{p=0}^\infty a_p(n) $$ with $$a_p(n)=\left\{\matrix{\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n}&\hbox{if}& 0\leq p<n/2\cr0&\hbox{otherwise}}\right.$$ Now, since $\Gamma(1)=1$ and $\Gamma'(1)=-\gamma$ we have, for a fixed $p$ and large $n$: $$(p-n)\ln\Gamma\left(1-\frac{p}{n}\right)=(p-n)\ln\left(1+\frac{\gamma p}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)=-\gamma p+\mathcal{O}\left(\frac{1}{n}\right)$$ Thus $$ \forall\,p\geq 0,\quad \lim_{n\to\infty}a_p(n)=e^{-\gamma p}.\tag{1} $$ Now, we will need the next lemma.
Lemma. For $t\in[1/2,1]$ we have $(\Gamma(t))^{t/(1-t)}\geq \Gamma(1/2)=\sqrt{\pi}.$
Taking, this lemma for granted, we conclude by taking $t=1-p/n$ when $0\leq p<n/2$, that $$ \forall\,p\geq 0,n\geq 1,\quad a_p(n)\leq \left(\frac{1}{\sqrt{\pi}}\right)^p. \tag{2} $$ and clearly, $$\sum_{p=0}^\infty \left(\frac{1}{\sqrt{\pi}}\right)^p<+\infty\tag{3}$$ Combining $(1)$, $(2)$ and $(3)$ we conclude that $$ \lim_{n\to\infty}I_3(n)=\lim_{n\to\infty}\sum_{p=0}^\infty a_p(n) =\sum_{p=0}^\infty\lim_{n\to\infty}a_p(n)= \sum_{p=0}^\infty e^{-\gamma p}=\frac{e^\gamma}{e^\gamma-1}.$$
The desired conclusion follows: $$ \lim_{n\to\infty}\sum_{1\leq k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}= \lim_{n\to\infty}(I_1(n)+I_2(n)+I_3(n))=\frac{e^\gamma}{e^\gamma-1}. $$
Proof of the Lemma. Let $f(t)=\dfrac{t}{1-t}\ln\Gamma(t)$. Then $f'(t)=\dfrac{g(t)}{(1-t)^2}$ with $$g(t)=\ln\Gamma(t)+t(1-t)\psi(t);\quad\hbox{where $\psi(t)=\Gamma'(t)/\Gamma(t)$}$$ and $g'(t)=(1-t)h(t)$ with $$h(t)=2\psi(t)+t\psi'(t)$$ and finally $h'(t)=3\psi'(t)+t\psi''(t)=\sum_{k=0}^\infty\frac{3k+t}{(k+t)^3}>0$.
So, $h$ is increasing, and $\lim_{t\to0^+}h(t)=-\infty$, $h(1)=\frac{\pi^2}{6}-2\gamma>0$. This proves that $h(t)<0$ for $0<t<x_0$ and $h(t)>0$ for $x_0<t<1$, for some $x_0$.
And $g$ is decreasing on $[0,x_0]$ and increasing on $[x_0,1]$. But $\lim_{t\to0^+}g(t)=+\infty$, $g(1)=0$. This proves that $g$ has exactly one change of sign on $(0,1)$ from positive to negative. This proves that the minimum of $f$ on $[1/2,1]$ is $\min(f(1/2),f(1))=f(1/2)$, and the lemma is proved.
Since $\log(\Gamma(x))$ is convex, $\log(\Gamma(x))\ge-\gamma(x-1)$ and $\log(\Gamma(x))\ge(1-\gamma)(x-2)$. Thus, $\log(\Gamma(x))\ge-\gamma+\gamma^2$. That is, $\Gamma(x)\ge e^{-\gamma+\gamma^2}=0.78345806514\gt\frac34$.
For $1\le k\le\frac23n$, $$ \begin{align} \Gamma\left(\frac kn\right)^{-k} &=\left(\frac kn\right)^k\,\Gamma\left(1+\frac kn\right)^{-k}\\ &\le\left(\frac{4k}{3n}\right)^k\tag1 \end{align} $$ Since $\left(1+\frac1k\right)^k\lt e\lt\left(1+\frac1k\right)^{k+1}$, we have $$ \frac4{3n}ke\le\frac4{3n}\frac{(k+1)^{k+1}}{k^k}\le\frac4{3n}(k+1)e\tag2 $$ Therefore, $\left(\frac{4k}{3n}\right)^k$ decreases while $\frac kn\lt\frac{3}{4e}$, then it increases.
Thus, applying $(1)$ and $(2)$, $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^{2n/3}\Gamma\left(\frac kn\right)^{-k} &\le\lim_{n\to\infty}\overbrace{\ \ \ \ \frac4{3n}\ \ \ \ \vphantom{\left(\frac89\right)^{\!\frac23n}}}^{k=1}+\overbrace{\frac23n\max\!\left(\frac{64}{9n^2},\left(\frac89\right)^{\!\frac23n}\right)}^{2\le k\le\frac23n}\\ &=0\tag3 \end{align} $$
For $0\le k\le\frac13n$, since $\Gamma\left(\frac{n-k}n\right)\ge1+\frac{\gamma k}n$ and $\left(1+\frac{\gamma k}n\right)^{n+\gamma k}\ge e^{\gamma k}$, $$ \begin{align} \Gamma\left(\frac{n-k}n\right)^{k-n} &\le\left(1+\frac{\gamma k}n\right)^{k-n}\\[3pt] &\le e^{-\gamma k\frac{n-k}{n+\gamma k}}\\[9pt] &\le e^{-\frac{2\gamma}{3+\gamma}k}\tag4 \end{align} $$ Furthermore, since $\Gamma\left(\frac{n-k}n\right)=1+\frac{\gamma k}n+O\left(\frac kn\right)^2$, $$ \lim_{n\to\infty}\Gamma\left(\frac{n-k}n\right)^{k-n}=e^{-\gamma k}\tag5 $$ by Dominated Convergence, $(4)$ and $(5)$ show that $$ \begin{align} \lim_{n\to\infty}\sum_{k=2n/3}^n\Gamma\left(\frac kn\right)^{-k} &=\lim_{n\to\infty}\sum_{k=0}^{n/3}\Gamma\left(\frac{n-k}n\right)^{k-n}\\ &=\sum_{k=0}^\infty e^{-\gamma k}\\[3pt] &=\frac{e^\gamma}{e^\gamma-1}\tag6 \end{align} $$
Putting $(3)$ and $(6)$ together, we get $$ \lim_{n\to\infty}\sum_{k=1}^n\Gamma\left(\frac kn\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}\tag7 $$