Straight line complexity of monomials
If $$f=(\Sigma_{i=1}^n x_i)^{2^n}$$
then it has ${2^n+n-1\choose n-1} \approx 2^{n^2}$ monomials and $L(f)=O(n)$.
By a counting argument, there are $2^{O(n\log n)}$ straight-line programs of length $O(n)$.
As $f$ has more monomials, for some we need a longer program.
In fact this argument gives a monomial $m$ for which $L(m)=\widetilde\Omega(L^2(f))$.