Let $1= H_0\lhd H_1\lhd\cdots\lhd H_n=G$. Assume that $H_i/H_{i-1}$ is nonabelian. Show that there exists $\chi\in Irr(G)$ with $\chi(1)\ge 2^n$.
Hint Use Corollary 6.17(Gallagher): Let $N\lhd G$ and let $\chi\in Irr(G)$ be such that $\chi|_N=\theta\in Irr(N)$. Then the characters $\beta\chi$ for $\beta\in Irr(G/N)$ are irreducible, distinct for distinct $\beta$ and are all of the irreducible constituents of $\theta^G$.
My attempt:
We can argue by induction on the length $r$ of the subnormal series. The base case is clear. Suppose it is true for all $r<n$, then for $r=n$, we notice that $H_n/H_{n-1}$ is non abelian implies there exists an irreducible character $\beta\in Irr(H_n/H_{n-1})$ such that $\beta(1)\ge 2$. By the induction hypothesis, there exists a $\psi\in Irr(H_{n-1})$ with $\psi(1)\ge 2^{n-1}$. To use Corollary 6.17, it suffices to show that $\psi$ is extendable to $H_n$.
Then I got stuck... Any hint?
I figured it out. With the same notations in the question we have the following:
Consider $\psi^G$. Note that $\psi^{G}(1)=|G:H_{n-1}|\psi(1)\ge 2^{n}$.
Prove by contradiction. Suppose any irreducible character of $G$ has degree less than $2^{n}$.
Let $\gamma$ be an irreducible constituent of $\psi^G$, that is $[\psi^G,\gamma]_{G}\ge 1$, by the Frobenius reciprocity, we have $$[\psi,\gamma|_{H_{n-1}}\ ]_{H_{n-1}}=[\psi^G,\gamma]_{G}\ge 1. $$
Clifford's theorem gives us $\gamma|_{H_{n-1}}=e\sum_{i=1}^t\psi^{(i)}$ where $e=[\psi^G,\gamma]_G$ and $\psi^{(i)}$ denotes the conjugate of $\psi$.
So $\gamma(1)=et\psi(1)\ge et2^{n-1}$. But $\gamma(1)<2^n$. So $e=t=1$. Thus $\gamma|_{H_{n-1}}=\psi$. And we know that $\psi$ is extendible to $G$. Hence $\gamma\beta(1)\ge 2^{n}$ and $\gamma\beta\in Irr(G)$ by Corollary 6.17 which is a contradiction.