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,
Look at the sublist,
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.
*)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.