Prove that $\lim_{k\to\infty}{ \sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}} - k } = \frac{\sum_{i=1}^{n}{x_i}}{n}$

Assume that $k$ is greater than twice $|x_1|+\ldots+|x_n|$ and also greater than twice $M=x_1^2+\ldots+x_n^2$. Over the interval $\left(-\frac{1}{2},\frac{1}{2}\right)$ we have $$ e^x = 1+x+C(x)x^2, \qquad \log(1+x)=x+D(x)x^2 $$ with $|C(x)|,|D(x)|\leq 1$. It follows that

$$\begin{eqnarray*}\text{GM}(x_1+k,\ldots,x_n+k)&=&k\cdot \text{GM}\left(1+\tfrac{x_1}{k},\ldots,1+\tfrac{x_n}{k}\right)\\&=&k\exp\left[\frac{1}{n}\sum_{j=1}^{n}\log\left(1+\frac{x_j}{k}\right)\right]\\&=&k\exp\left[\frac{1}{k}\sum_{j=1}^{n}\frac{x_j}{n}+\Theta\left(\frac{M}{nk^2}\right)\right]\\&=&k\left[1+\frac{1}{k}\sum_{j=1}^{n}\frac{x_j}{n}+\Theta\left(\frac{M}{nk^2}\right)\right]\end{eqnarray*}$$ and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $\frac{\text{Var}(x_1,\ldots,x_n)}{\text{AM}(x_1,\ldots,x_n)}$. A translation towards the right leaves the variance unchanged and increases the mean.

You could even get an interesting asymptotics considering $$y_n=\sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}}$$ $$ \log(y_n)=\frac 1n \sum_{i=1}^{n}\log\left({(x_i+k)}\right)=\frac 1n \sum_{i=1}^{n}\left(\log(k)+\log\left(1+\frac {x_i}k\right)\right)$$ Assuming that $x_1<x_2<\cdots < x_n \ll k$, use the Taylor expansion of $\log\left(1+\frac {x_i}k\right)$. Then continuing with Taylor series $$y_n=e^{\log(y_n)}\implies y_n=k+\frac{\sum_{i=1}^{n} x_i}{n}+\frac{\left(\sum_{i=1}^{n} x_i\right)^2-n\sum_{i=1}^{n} x^2_i}{2n^2 k}+O\left(\frac{1}{k^2}\right)$$ making $$ \sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}} - k =\frac{\sum_{i=1}^{n} x_i}{n}+\frac{\left(\sum_{i=1}^{n} x_i\right)^2-n\sum_{i=1}^{n} x^2_i}{2n^2 k}+O\left(\frac{1}{k^2}\right)$$

For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $\frac{2572671}{200000}\approx 12.8634$ while the exact calculation would lead to $\approx 12.8639$

Your expression under limit can be written as $k(a-b) $ where $$a=\sqrt[n]{\prod_{i=1}^{n}\left(1+\frac{x_i}{k}\right)}, b=1$$ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $$c=a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}$$ tends to $n$ as $k\to\infty$. And $$a^n-b^n=\prod_{i=1}^{n}\left(1+\frac{x_i}{k}\right)-1=\frac{1}{k}\sum_{i=1}^{n}x_i+o(1/k)$$ and thus $$k(a-b) =\frac{1}{n}\sum_{i=1}^{n}x_i+o(1)$$ and the desired limit as $k\to\infty$ is $\dfrac{1}{n}\sum\limits_{i=1}^{n}x_i$.