Summation of an expression

REVISED ANSWER.

In retrospect, deriving the approximation is quite easy. Indeed, $$ \sum\limits_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} = e\sum\limits_{j = 1}^n {e^{ - 1} \frac{{j^{k + 1} }}{{j!}}} \approx e\sum\limits_{j = 0}^\infty {e^{ - 1} \frac{{j^{k + 1} }}{{j!}}} = eB_{k + 1}. $$ (For large $k$, you may consider Asymptotic limit and bounds.)

ORIGINAL ANSWER.

Assume that $n$ is sufficiently large. Since $$ \sum\limits_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} = \sum\limits_{j = 0}^{n - 1} {\frac{{(j + 1)^k }}{{j!}}} , $$ the problem reduces to approximating $\sum\nolimits_{j = 0}^{n - 1} {\frac{{j^m }}{{j!}}} $, for $0 \leq m \leq k$. Now, $e^{ - 1} \sum\nolimits_{j = 0}^\infty {\frac{{j^m }}{{j!}}} $ is the $m$-th moment of the Poisson distribution with mean $1$. For the latter, see, for example, this.

EDIT: Specifically, the approximation (with the right-hand side being an upper bound) is $$ \sum\limits_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} \approx e\sum\limits_{m = 0}^k {{k \choose m}B_m }, $$ where $B_m$ is the $m$-th Bell number (a list is given here). Numerical results indicate that this approximation is very accurate, even for moderate values of $n$. For example, the absolute error for $n=15$, $k=3$ is $\approx 3.4 \times 10^{-9}$; for $n=20$, $k=5$ is $\approx 1.8 \times 10^{-12}$; for $n=20$, $k=7$ is $\approx 7.9 \times 10^{-10}$; for $n=23$, $k=9$ is $\approx 5.8 \times 10^{-11}$.

EDIT: If only the relative error is concerned, then the approximation is quite accurate even for relatively small $n$ values (of course, depending on $k$). For example, for $n=8$, $k=4$: $\approx 141.156$ compared to $\approx 141.351$; for $n=9$, $k=5$: $\approx 551.484$ compared to $\approx 551.811$; for $n=10$, $k=6$: $\approx 2383.359$ compared to $\approx 2383.933$.

EDIT: As Mike Spivey observed, using the identity $$ \sum\limits_{m = 0}^k {{k \choose m}B_m } = B_{k + 1}, $$ the above approximation can be simplified greatly to $$ \sum\limits_{j = 1}^n {\frac{{j^k }}{{(j - 1)!}}} \approx e B_{k+1}. $$

There might be something in Hall and Knight, Higher Algebra. On page 333, they have the formula, $\sum_1^{n+2}{j^2+j-1\over(j+2)!}={1\over2}-{1\over(n+2)(n!)}$.

The infinite sum is a multiple of $e\ $ and a Bell number. So find the discrepancy (which will be very small), maybe on order of $e^{-2n}$).