Question on Proof by Induction for formula of Bernoulli numbers

Question

I am trying to prove by induction that $F_{n-1}$ = $\frac{-1}{n}\sum_{a=0}^{n-2}{n \choose a} F_a$. The sequence $F_n$ is the Bernoulli numbers, with $F_0=1, F_2=-1/2$ etc. so the formula in question is a recurrence formula for the Bernoulli numbers. I understand how to take the base case, but am struggling with making the inductive step. Should it be something like this: Base Case: $F_{n-1}$ = $\frac{-1}{n}\sum_{a=0}^{n-3}{n \choose a} F_a$ Inductive Step: $F_{n} = \frac{-1}{n+1}(\sum_{a=0}^{n-1}{n\choose a}F_a + {n\choose n-1}F_{n-1})$. I'm then confused about where to go because of the $\frac{-1}{n+1}$ out in front. Do I try to modify it by algebra to look like $F_{n} = \frac{-1}{n+1}(\sum_{a=0}^{n-1}{{n+1}\choose a}F_a$ ? Please help!