Inverse of the sum $\sum\limits_{j=1}^k (-1)^{k-j}\binom{k}{j} j^{\,k} a_j$ $k\in\mathbb{N}$ 
The inverse of the sum $$b_k:=\sum\limits_{j=1}^k (-1)^{k-j}\binom{k}{j} j^{\,k} a_j$$ is obviously 
$$a_k=\sum\limits_{j=1}^k \binom{k-1}{j-1}\frac{b_j}{k^j}$$ . 
How can one proof it (in a clear manner)? 
Thanks in advance.

Background of the question: 
It’s  $$\sum\limits_{k=1}^\infty \frac{b_k}{k!}\int\limits_0^\infty \left(\frac{t}{e^t-1}\right)^k dt =\sum\limits_{k=1}^\infty \frac{a_k}{k}$$  with $\,\displaystyle b_k:=\sum\limits_{j=1}^k (-1)^{k-j}\binom{k}{j}j^{\,k}a_j $. 
Note:
A special case is $\displaystyle a_k:=\frac{1}{k^n}$ with $n\in\mathbb{N}$ and therefore $\,\displaystyle b_k=\sum\limits_{j=1}^k (-1)^{k-j}\binom{k}{j}j^{\,k-n}$ (see Stirling numbers of the second kind) 
$$\sum\limits_{k=1}^n \frac{b_k}{k!}\int\limits_0^\infty \left(\frac{t}{e^t-1}\right)^k dt =\zeta(n+1)$$ and the invers equation can be found in A formula for $\int\limits_0^\infty (\frac{x}{e^x-1})^n dx$ . 

In this proof, the binomial identity 
$$\binom{m}{n}\,\binom{n}{s}=\binom{m}{s}\,\binom{m-s}{n-s}$$
for all integers $m,n,s$ with $0\leq s\leq n\leq m$ is used frequently, without being specifically mentioned.  A particular case of importance is when $s=1$, where it is given by
$$n\,\binom{m}{n}=m\,\binom{m-1}{n-1}\,.$$
First, rewrite 
$$b_k=k\,\sum_{j=1}^{k}\,(-1)^{k-j}\,\binom{k-1}{j-1}\,j^{k-1}\,a_j\,.$$
Then,
$$\sum_{k=1}^l\,\binom{l-1}{k-1}\,\frac{b_k}{l^k}=\sum_{k=1}^l\,\binom{l-1}{k-1}\,\frac{k}{l^k}\,\sum_{j=1}^k\,(-1)^{k-j}\,\binom{k-1}{j-1}\,j^{k-1}\,a_j\,.$$
Thus,
$$\begin{align}
\sum_{k=1}^l\,\binom{l-1}{k-1}\,\frac{b_k}{l^k}&=\sum_{j=1}^l\,\frac{a_j}{j}\,\sum_{k=j}^l\,(-1)^{k-j}\,\binom{l-1}{k-1}\,\binom{k-1}{j-1}\,k\left(\frac{j}{l}\right)^k
\\
&=\sum_{j=1}^l\,\frac{a_j}{j}\,\binom{l-1}{j-1}\,\sum_{k=j}^l\,(-1)^{k-j}\,\binom{l-j}{k-j}\,k\left(\frac{j}{l}\right)^k\,.
\end{align}$$
Let $r:=k-j$.  We have
$$\sum_{k=1}^l\,\binom{l-1}{k-1}\,\frac{b_k}{l^k}=\sum_{j=1}^l\,\frac{a_j}{j}\,\binom{l-1}{j-1}\,\left(\frac{j}{l}\right)^j\,\sum_{r=0}^{l-j}\,(-1)^r\,\binom{l-j}{r}\,(r+j)\,\left(\frac{j}{l}\right)^{r}\,.\tag{*}$$
Now, if $j=l$, then 
$$\sum_{r=0}^{l-j}\,(-1)^r\,\binom{l-j}{r}\,(r+j)\,\left(\frac{j}{l}\right)^{r}=l\,.$$
If $j<l$, then
$$\begin{align}
\sum_{r=0}^{l-j}\,(-1)^r\,\binom{l-j}{r}\,r\,\left(\frac{j}{l}\right)^{r}&=-(l-j)\left(\frac{j}{l}\right)\,\sum_{r=1}^{l-j}\,(-1)^{r-1}\,\binom{l-j-1}{r-1}\,\left(\frac{j}{l}\right)^{r-1}
\\&=-j\left(1-\frac{j}{l}\right)\,\left(1-\frac{j}{l}\right)^{l-j-1}=-j\left(1-\frac{j}{l}\right)^{l-j}
\end{align}$$
and
$$\sum_{r=0}^{l-j}\,(-1)^r\,\binom{l-j}{r}\,j\left(\frac{j}{l}\right)^r=j\left(1-\frac{j}{l}\right)^{l-j}\,.$$
Consequently,
$$\sum_{r=0}^{l-j}\,(-1)^r\,\binom{l-j}{r}\,(r+j)\,\left(\frac{j}{l}\right)^{r}=\begin{cases}
0\,,&\text{if }j<l\,,\\
l\,,&\text{if }j=l\,.
\end{cases}$$
From (*),
$$\sum_{k=1}^l\,\binom{l-1}{k-1}\,\frac{b_k}{l^k}=\frac{a_l}{l}\,\binom{l-1}{l-1}\,\left(\frac{l}{l}\right)^l\,l=a_l\,.$$
