Yes. I will prove a lemma using induction, which I will apply to prove the result. I will use the convention $[a, b]=a^{-1}b^{-1}ab$ in my proofs: using a different convention does not alter the result, and the proofs would be equivalent up to notation changes.
Lemma: If $a$ and $b$ have order two then the following holds.
If $n$ is even, $[a,\underbrace{b,\ldots,b}_n]=(ba)^{2^n}$
If $n$ is odd, $[a,\underbrace{b,\ldots,b}_n]=(ab)^{2^n}$
Proof: We will induct on $n$. If $n=1$ then $[a, b]=a^{-1}b^{-1}ab=abab=(ab)^2$, so the result holds.
There are two cases for the induction step: $n=k-1$ is even, or $n=k-1$ is odd. Suppose the result holds for $n=k-1$ where $n=k-1$ is odd. We will prove it for $k$. So, begin with the following.
$$\begin{align*}
[a,\underbrace{b,\ldots,b}_k]
&=[a,\underbrace{b,\ldots,b}_{k-1}, b]\\
&=[(ab)^{2^{k-1}}, b]\\
&=(ab)^{-2^{k-1}}b(ab)^{2^{k-1}}b\\
&=(ba)^{2^{k-1}}(ba)^{2^{k-1}}\\
&=(ba)^{2^{k-1}+2^{k-1}}=(ba)^{2^{k}}
\end{align*}$$
The proof for $n=k-1$ odd is analogous.
Theorem: In a group of exponent $2^n$, the following identity holds.
$$[x^{2^{n-1}},\underbrace{y^{2^{n-1}},\ldots,y^{2^{n-1}}}_n]=1$$
Proof: From the above lemma, we see that $[x^{2^{n-1}},\underbrace{y^{2^{n-1}},\ldots,y^{2^{n-1}}}_n]$ is equal to either $(x^{2^{n-1}}y^{2^{n-1}})^{2^n}$ or $(y^{2^{n-1}}x^{2^{n-1}})^{2^n}$. As the group has exponent $2^n$, the result follows.
