The coefficient of $x^k$ in $b_r(x)$ for $k=1,2,\ldots,r$ is $\binom{r}{k}B_{r-k}$

The proof is by induction on $r$ For $r=1$ note that $b_1(x)=x-\frac{1}{2}$ so that $[x]b_1(x)=1=\binom{1}{1}B_0$

Writing $[x^k]f(x)$ for the coefficient of $x^k$ in a polynomial $f(x)$ note that for $k=1,2,\ldots,r$ $$[x^k]b_r(x)=\frac{1}{k}[x^{k-1}]b_r'(x)=\frac{r}{k}[x^{k-1}]b_{r-1}(x)$$ since $b_r'(x)=rb_{r-1}(x)$ By induction, $$\frac{r}{k}[x^{k-1}]b_{r-1}(x)=\frac{r}{k}\binom{r-1}{k-1}B_{r-k}=\binom{r}{k}B_{r-k}$$

Thus the Bernoulli polynomials can be written $$b_r(x) = \sum_{k=1}^r \binom{r}{k}B_{r-k}x^k + B_r$$