Where was the mistake

Question

we know that $$\frac{\pi^2}{6}=\int_{0}^{\infty}\frac{t}{e^t-1}dt$$, we also know that $\frac{t}{e^t-1}$ is the generating function for the Bernoulli numbers i.e $ \frac{t}{e^t-1} =\sum_{n=1}^{\infty}\frac{B_nt^n}{n!}$. If we use this generating function in the above i.e $$\frac{\pi^2}{6}=\int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{B_nt^n}{n!}dt$$ changing the order of integration $$\frac{\pi^2}{6}=\sum_{n=1}^{\infty}B_n\int_{0}^{\infty}\frac{t^n}{n!}dt$$ but this is not true. Where is the flaw.