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$\newcommand\Ga\Gamma$ Note that $\Ga(x)=\Ga(1+x)/x$ for $x>0$ and $-n=1-\int_0^1 x^{-n/(n+1)}\,dx$ for $n>0$.

So, the limit in question is $$1+\lim_n J_n,$$ where $$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ $$f_n(x):=g(x)-h_n(x),$$ $$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all $n\ge1$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is $$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$

$\newcommand\Ga\Gamma$ Note that $\Ga(x)=\Ga(1+x)/x$ for $x>0$ and $-n=1-\int_0^1 x^{-n/(n+1)}\,dx$ for $n>0$.

So, the limit in question is $$1+\lim_n J_n,$$ where $$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ $$f_n(x):=g(x)-h_n(x),$$ $$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all $n\ge1$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is $$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$

As seen from the proof, the rate of convergence here is $O(1/n)$. So, the limit value $0.75330\ldots$ is in agreement with the value of the integral you computed for $n=100$.

$\newcommand\Ga\Gamma$ The limit in question is $$1+\lim_n J_n,$$ where $$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ $$f_n(x):=g(x)-h_n(x),$$ $$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all natural $n$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is $$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$

$\newcommand\Ga\Gamma$ Note that $\Ga(x)=\Ga(1+x)/x$ for $x>0$ and $-n=1-\int_0^1 x^{-n/(n+1)}\,dx$ for $n>0$.

So, the limit in question is $$1+\lim_n J_n,$$ where $$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ $$f_n(x):=g(x)-h_n(x),$$ $$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all $n\ge1$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is $$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$

$\newcommand\Ga\Gamma$ The limit in question is $$1+\lim_n J_n,$$ where $$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ $$f_n(x):=g(x)-h_n(x),$$ $$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all natural $n$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}n_n(x)\,dx\to0$. Thus, the limit in question is $$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$

$\newcommand\Ga\Gamma$ The limit in question is $$1+\lim_n J_n,$$ where $$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ $$f_n(x):=g(x)-h_n(x),$$ $$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all natural $n$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is $$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$