PLONK Prod Check Proof - why does it have to be proven upto the last element of the set? It should be enough to prove it upto last but one element

I am going through Dan Boneh's video on PLONK - https://www.youtube.com/watch?v=LbpPCN-f_XA&t=952s

At around 19 minutes, he gets to the Prod Check Gadget.

Background:

$\omega \in \mathbb F_p$ is the primitive $k$th root of unity (i.e. $\omega^{k} = 1$)

$\Omega = \{1, \omega, \omega^{2}, ..., \omega^{k-1}\}$

Let $t(1) = f(1)$ and $t(\omega^s) = \prod_{i=0}^s f(\omega^i)$ for $s = 1, ..., k-1$.

Prover has to prove that:

$\prod_{a \in \Omega} f(a) = 1$

Boneh says that Prover can prove the above by proving 2 things

1) $t(\omega^{k-1}) = 1$

and

2) $t(\omega\cdot x) - t(x)\cdot f(\omega \cdot x) = 0$ for all $x \in \Omega$ (including at $x = \omega^{k-1})$

My question is about the including in the 2nd proof.

I think it needs to be proved only up to $x= \omega^{k-2}$ & the last element $\omega^{k-1}$can be ignored.

If we take $x= \omega^{k-1}$, then 2nd equation

$t(\omega\cdot x) - t(x)\cdot f(\omega \cdot x) \stackrel {?}{=} 0$

becomes

$t(\omega \cdot \omega^{k-1}) - t(\omega^{k-1})\cdot f(\omega \cdot \omega^{k-1}) \stackrel {?}{=} 0$

i.e.

$t(\omega^k) - t(\omega^{k-1})\cdot f(\omega^k) \stackrel {?}{=} 0$

I don't think it's needed to prove this at all, since the final equality we want to prove is $t(\omega^{k-1}) = 1$ - we never have to go beyond the $x = k-2$th power of $\omega$ (i.e. $x = \omega^{k-2}$)

Using $x = \omega^{k-2}$ proves

$t(\omega^{k-1}) - t(\omega^{k-2})\cdot f(\omega^{k-1}) = 0$

& we have already proved $\omega^{k-1} = 1$

So there is no need for $x = \omega^{k-1}$.

So why is Boneh very particularly saying (including at $x = \omega^{k-1})$?

What am I missing?