I found an answer:
We will need the following easy to prove (I can expand here if necessary)
relations:
\begin{eqnarray*}
B_{k+1}(x) = \sum_{i=0}^{k+1} B_i
\binom{k+1}{i}
x^{k+1-i} \quad \quad (1),
\end{eqnarray*}
and
\begin{eqnarray*}
\sum_{i=0}^{n-1} (i+m)^{k-1} =
\frac{B_{k}(n+m) - B_{k}(m)}{k} \quad \quad (2)
\end{eqnarray*}
Let us start:
\begin{eqnarray*}
\sum_{k=0}^{\infty} \frac{1}{(k+x)^s} =
\sum_{k=0}^{N-1} \frac{1}{(k+x)^s} +
\sum_{k=N}^{\infty} \frac{1}{(k+x)^s}
\end{eqnarray*}
and, in the second sum, use the Euler-Maclaurin series with
$h=1, a=N, n= \infty$, $f(k)=1/(k+x)^s$. Then
\begin{eqnarray*}
\sum_{i=0}^{n-1} f(a + i h) h = \int_a^b f(t) dt +
\left . \sum_{k=1}^{m} \frac{B_k
h^k}{k!} f^{(k-1)} (t) \right |_a^b + R_m \\
\sum_{i=0}^{\infty} \frac 1{(N + i+x)^s} =
\int_{N}^{\infty} \frac {dt}{(t+x)^s} + \left .
\sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{\Gamma(s+k-1)}{\Gamma(s) (t+x)^{s+k-1}}
\right |_{N}^{\infty}+ R_m\\
\sum_{k={N}}^{\infty} \frac 1{(k+x)^s} = \frac {1}{(s-1)(N+x)^{s-1}} -
\sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{\Gamma(s+k-1)}{\Gamma(s) \,
(N+x)^{s+k-1}}+ R_m
\end{eqnarray*}
with
\begin{eqnarray*}
R_m = \int_N^{\infty} B_m \left \{ \frac{t-a}{h} \right \}
\frac{\Gamma(s+m)}{ \Gamma(s)} \frac{1}{ (t+x)^{s+m+1}}.
\end{eqnarray*}
Then
\begin{eqnarray*}
\zeta(s,x)= \sum_{k=0}^{N-1} \frac 1{(k+x)^s} +
\frac {1}{(s-1)\,(N+x)^{s-1}} -
\sum_{k=1}^{m} \frac{B_k}{k!} \frac{\Gamma(s+k-1)}{\Gamma(s) (N+x)^{s+k-1}}+ R_m,
\end{eqnarray*}
and taking the limit as $m \to \infty$ (since the
sum converges and $R_m \to 0$, this is easy to show for this function), we find
\begin{eqnarray*}
\zeta(s,x) = \sum_{k=0}^{N-1} \frac{1}{(k+x)^s} +
\frac{1}{ (s-1) (N+x)^{s-1}} + \sum_{k=1}^{\infty} \frac{B_k }{k!}
\frac{\Gamma(s+k-1)}{\Gamma(s)}.
\end{eqnarray*}
We want to choose $s=1-j$, for $j$ a positive integer. First,
we consider the
Pochhammer Symbold
\begin{eqnarray*}
(s)_{k-1} = \frac{ \Gamma(s + k -1)}{\Gamma(s)}.
\end{eqnarray*}
We prove the reflection formula for the Pochhammer symbol,
\begin{eqnarray*}
(-t)_n = (-1)^n (t -n + 1)_n
\end{eqnarray*}
This is,
\begin{eqnarray*}
(-t)_n &=& \frac{ \Gamma(-t+n)}{\Gamma(-t)} \\
&=& \frac{ (-t+n-1)(-t+n-2) \cdots (-t) \cancel{\Gamma(-t)}}{\cancel{\Gamma(-t)}} \\
&=& (-1)^n (t-n+1)(t-n+2) \cdots t \\
&=& (-1)^n t (t-1) \cdots (t-n+1) \\
&=& (-1)^n \frac{\Gamma(t+1)}{\Gamma(t-n+1)} \\
&=& (-1)^n (t-n+1)_n
\end{eqnarray*}
Then we find that for $s=1-j$
\begin{eqnarray*}
\frac{\Gamma(s+k-1)}{k! \; \Gamma(s)} = \frac{(1-j)_{k-1}}{k!} =
(-1)^{k-1} \frac{(j-1-k+1 +1)_{k-1}}{k!}
= (-1)^{k-1} \frac{ (j-k+1)_{k-1}}{k!}
= (-1)^{k-1} \frac{ (j-1)!}{k! \, \Gamma(j-k)}
\end{eqnarray*}
and we can write the formula for $\zeta(1-j,x)$ as
\begin{eqnarray*}
\zeta(1-j,x) &=&
\sum_{k=0}^{N-1} (k+x)^{j-1}
- \frac{ (N+x)^j}{j}
+ \frac{1}{j} \sum_{k=1}^j (-1)^{k-1} B_k \;
(N+x)^{j-k} \binom{j}{k} .
\\
&=&
\sum_{k=0}^{N-1} (k+x)^{j-1}
- \frac{1}{j} \sum_{k=0}^j (-1)^{k} B_k \;
(N+x)^{j-k} \binom{j}{k} .
\end{eqnarray*}
where we introduced the independent term into the second sum. We will change
the alternating sign $(-1)^{k-1}$ for a $-1$ sign, since for $k > 1$, odd $B_{k}=0$, and the
first two terms (indices $k=0,1$) have the right sign.
We recognize, from equation (1), into the equation above that
\begin{eqnarray*}
\zeta(1-j,x) =
\sum_{k=0}^{N-1} (k+x)^{j-1}
- \frac{1}{j} \, B_{j}(N+x).
\end{eqnarray*}
where we removed the alternating sign $(-1)^{k-1}$ since for $i>1$ odd $B_k=0$.
We now use equation (2) to write
\begin{eqnarray*}
\zeta(1-j,x) =
\frac{B_{j}(N+x) - B_{j}(x)}{j}
- \frac{1}{j} \, B_{j}(N+x)
= -\frac{B_j(x)}{j}.
\end{eqnarray*}
This is what we wanted to show.
