It is an easy proof. Since n is highly composite we have that t(n)>t(i) for all 1<=i<n.

Now we are told that t(m)>t(n).

Therefore m>n because otherwise m<=n contradicts the fact that n is highly composite.

Now consider the list of integers,

t(1),t(2),t(3),...,t(n),t(n+1),...,t(m)

Look at the sublist,

t(n+1),t(n+2),...,t(m)

And choose the smallest maximal element (possible because this is finite).*

Call it k, thus, n<k<=m.

By our choice of k we have that t(k)>=t(i) for all n+1<=i<=m.

But we also have that t(k)>=t(m)>t(n)>t(j) for all 1<=j<n.

Thus, t(k)>=t(j) for all 1<=j<=k

Thus, t(k)>t(j) for all 1<=j<k by our choice of smallest.

Thus, by definition k is highly composite.

Q.E.D.

*)Given a non-empty set S of positive integers a "maximal element" is M so that M>=x for all x in S.

By smallest I mean the first in the list.