(𝜆x. (𝜆y. y)) (𝜆a. (𝜆b.a)) beta reduction

I've came across an example and I'm not quite sure on how the solution was met after performing beta-reduction on the following expression. It doesn't show any of the steps. Any help is appreciated!

(𝜆x. (𝜆y. y)) (𝜆a. (𝜆b.a))

(𝜆y. y)

The $$(\beta)$$ rule tells us that an expression of the form $$(\lambda x . s)t$$ reduces to $$s[t/x]$$.
In your case we have $$t \equiv (\lambda a .(\lambda b. a))$$, and $$s \equiv \lambda y . y$$.
Then one $$\beta$$-reduction gives us $$s[t/x]$$, which is just equivalent to $$s$$ because $$x$$ doesn't appear free in $$s$$.