Vanishing of a certain Tor I am reading about the construction of the Affine Grassmannian in Dennis Gaitsgory's seminar notes
and there are some commutative algebra facts that I am not able to figure out by myself apparently, like the following:

Let $k$ be an algebraically closed field, $A$ a finite type $k$-algebra and $A\subseteq B$ some (non finite type) extension. Let $M$ be a finitely generated $A[[t]]$-module which is flat over $A$ (actually finite free over $A$) and $t$ acts nilpotently on $M$. Then, $\operatorname{Tor}^{A[[t]]}_n(M,B[[t]])=0$ for all $n>0$.

Intuitively, since we get $B[[t]]$ from $A[[t]]$ by extending only the coefficient ring $A$ in a "free" way, flatness over $A$ of $M$ should suffice, but I can't make it into a proof. Notice that $A[[t]]\otimes _A B\ne B[[t]]$ in general. Am I missing something obvious?
Edit:
After reading it again, it seems that the only additional hypothesis I missed is that $t$ acts nilpotently on $M$ which I think does not follow from what I have written. The relevant place in the notes is the first line on page 7 and a bit before that.
 A: Here's an attempt (could be mistakes so be wary!).
We have
$$
M \otimes_{A[[t]]}^{\mathbb{L}} B[[t]] \cong (M \otimes_{A[[t]]}^{\mathbb{L}} A[[t]]/t^n) \otimes_{A[[t]]}^{\mathbb{L}} B[[t]]. 
$$
By associativity this is (quasi-isomorphic) to 
$$
M \otimes_{A[[t]]}^{\mathbb{L}} B[[t]]/t^n.
$$
Since $A[[t]] \rightarrow A[[t]]/t^n$ is surjective, this is the same as
$$
M \otimes_{A[[t]]/t^n}^{\mathbb{L}} B[[t]]/t^n.$$
And this satisfies the conclusion you want because of your conditions on $M$. 
A: New attempt:
Lemma 24.6.6 in Vakil's AG notes states:

Suppose $N$ is an $R$-module and $t\in R$ is not a zero divisor on $N$. Then for any $R/(t)$-module $M$, we have 
  $$
\operatorname{Tor}_i^R(M,N)=\operatorname{Tor}_i^{R/(t)}(M,N/(t)).
$$
  (Actually it is stated with the roles of $M$ and $N$ reversed, but Tor is symmetric.)

Now, in our case, $R=A[[t]]$, $t^n$ is not a zero divisor on $N=B[[t]]$ and $M$ is actually an $A[[t]]/(t^n)$-module so we get
$$
\operatorname{Tor}_i^{A[[t]]}(M,B[[t]])=\operatorname{Tor}_i^{A[[t]]/(t^n)}(M,B[[t]]/(t^n)).
$$
Take a free resolution $F_{\bullet}\to B$ over $A$. Tensoring with $A[[t]]/(t^n)$ over $A$, we get
$$
F_{\bullet}\otimes _A A[[t]]/(t^n)\to B[[t]]/(t^n)
$$
Which is a free resolution of $B[[t]]/(t^n)$ over $A[[t]]/(t^n)$. Now, tensor with $M$ over $A[[t]]/(t^n)$, we get the complex
$$
(F_{\bullet}\otimes _A A[[t]]/(t^n)) \otimes _{A[[t]]/(t^n)} M \cong F_{\bullet}\otimes _A M
$$
On the on hand, its homology are precisely the Tor-s we want to calculate and on the other, it is exact by $A$-flatness of $M$, so its homology is zero.

Edit (final, I hope...): The second step is exactly the flat base change of Tor, but on the other module. The map $A\to A[[t]]/(t^n)$ is flat so we get
$$
\operatorname{Tor}_i^{A}(M,B)=\operatorname{Tor}_i^{A[[t]]/(t^n)}(M,B[[t]]/(t^n))
$$
Where we used $B\otimes _A A[[t]]/(t^n)=B[[t]]/(t^n)$ (for this, the reduction mod $t^n$ was necessary!) and now the LHS is zero by $A$-flatness of $M$. 
