[tex]\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}[/tex]
By the Stolz-Cesaro theorem, this limit exists if
[tex]\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}[/tex]
also exists, and the limits would be equal. The theorem requires that [tex]b_n[/tex] be strictly monotone and divergent, which is the case since [tex]k\in\mathbb N[/tex].
You have
[tex]a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}[/tex]
so we're left with computing
[tex]\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}[/tex]
This can be done with the help of Stirling's approximation, which says that for large [tex]n[/tex], [tex]n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n[/tex]. By this reasoning our limit is
[tex]\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}[/tex]
Let's examine this limit in parts. First,
[tex]\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}[/tex]
As [tex]n\to\infty[/tex], this term approaches 1.
Next,
[tex]\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}[/tex]
The term on the right approaches [tex]e^k[/tex], cancelling the [tex]e^{-k}[/tex]. So we're left with
[tex]\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}[/tex]
Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.
[tex]\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}[/tex]
Divide through the numerator and denominator by [tex]n^k[/tex]:
[tex]\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}[/tex]
So you can see that, by comparison, we have
[tex]\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}[/tex]
so this is the value of the limit.
