*Note:* This answer does not provide a closed expression but a *generating function* which might also be helpful for further calculations.

Let's consider OPs sum by exchanging for (my) convenience $i,j$ with $n,k$ and ignoring the factor $\ln(x)^n$.

\begin{align*}
\frac{1}{n!}&\sum_{k=0}^n\binom{n}{k}(n-k)^kx^k\tag{1}\\
&=\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}k^{n-k}x^{n-k}
\end{align*}

Let $A(z)$ denote the generating function of OPs expression (1). We show

The following is valid:

\begin{align*}
A(z) &:= \sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}k^{n-k}x^{n-k}\right)z^n\tag{2}\\
&=\exp\left(ze^{xz}\right)
\end{align*}

**Intermezzo:** *Bell polynomials*

According to Louis Comtet's *Advanced Combinatorics* section 3.3 ([3a']) the *partial Bell polynomials* $B_{n,k}=B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})$ are defined via

\begin{align*}
B_{n,k}=\frac{n!}{k!}[t^n]\left(\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)^k
\end{align*}

We use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ in the formal power series.

The specific case $B_{n,k}=B_{n,k}(1,2,\ldots,n-k+1)$ yields:

\begin{align*}
B_{n,k}&=\frac{n!}{k!}[t^n]\left(\sum_{m\geq 1}m\frac{t^m}{m!}\right)^k\\
&=\frac{n!}{k!}[t^n]\left(t\sum_{m\geq 0}\frac{t^{m}}{m!}\right)^k\\
&=\frac{n!}{k!}[t^{n}]t^k\exp(kt)\tag{3}\\
&=\frac{n!}{k!}[t^{n-k}]\sum_{m\geq 0}\frac{(kt)^{m}}{m!}\\
&=\frac{n!}{k!}\frac{k^{n-k}}{(n-k)!}\\
&=\binom{n}{k}k^{n-k}\tag{4}
\end{align*}

We observe the *partial Bell polynomials* $B_{n,k}(1,2,\ldots,n-k+1)=\binom{n}{k}k^{n-k}$ are the link between OPs expression and the path to finding the generating function $A(z)$.

Using the expression (3) in (1) we obtain

\begin{align*}
\frac{1}{n!}\sum_{k=0}^n&\binom{n}{k}k^{n-k}x^{n-k}\\
&=\sum_{k=0}^n\frac{1}{k!}[t^{n-k}]e^{kt}x^{n-k}\\
&=x^n[t^n]\sum_{k=0}^{\infty}\left(\frac{te^t}{x}\right)^k\frac{1}{k!}\tag{4}\\
&=x^n[t^n]\exp\left(\frac{te^t}{x}\right)\tag{5}
\end{align*}

In (4) we changed the upper limit of the index $k$ from $n$ to $\infty$ which does not contribute anything (just adding $0$'s).

From the last expression (5) we finally get

\begin{align*}
A(z) &=\sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}k^{n-k}x^{n-k}\right)z^n\\
&=\sum_{n=0}^{\infty}\left(x^n[t^n]\exp\left(\frac{te^t}{x}\right)\right)z^n\tag{6}\\
&=\sum_{n=0}^{\infty}\left([t^n]\exp\left(\frac{te^t}{x}\right)\right)(xz)^n\\
&=\exp\left(\frac{xze^{xz}}{x}\right)\tag{7}\\
&=\exp\left(ze^{xz}\right)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}

and the claim follows.

*Comment:*

$$A(z)=\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\left([t^n]A(t)\right)z^n$$