How to represent characteristic polynomial in terms of those of invariant direct sum subspaces? 
Suppose $V$ is a complex vector space and $V_1,...,V_m$ are nonzero subspaces of $V$ such that $V = V_1 \oplus ... \oplus V_m$. Suppose $T \in \mathcal{L}(V)$ and each $V_j$ is invariant under $T$. For each $j$, let $p_j$ denote the characteristic polynomial of $T|_{V_j}$. Prove that the characteristic polynomial of $T$ equals $p_1...p_m$.

The above question is from Linear Algebra Done Right 3rd ed. Chapter 8C problem 20.
I found and understood the proof from here. However since the book above I am (self) studying doesn't introduce determinant till the last chapter, I would be very grateful if someone could help me with a determinant free proof.
Note:
In the book the characteristic polynomial is defined as the product
$$(z−\lambda_1)^{m_1}⋯(z−\lambda_n)^{m_n}$$
for some $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional complex vector space, where $\lambda_k$ are the distinct eigenvalues of $T$ and $m_k$ it algebraic multiplicities.
 A: From the decomposition
$$
V=\bigoplus_{j=1}^m V_j\tag1
$$
define a basis $B_j$ for each $V_j$. Then $B:=\bigcup_{j=1}^m B_j$ is a basis of $V$.
Because $p_j$ is the characteristic polynomial of $T|_{V_j}$ and $V$ is complex then $\deg p_j=\dim V_j$, thus $\deg (\prod_{j=1}^m p_j)=\dim V$.
Now if $p_j(T)v=0$ then clearly $p(T)v=0$, where we define $p:=\prod_{j=1}^m p_j$. Then is easy to see that for each $v_k\in B$ we have that $p(T)v_k=0$, hence $p(T)=0$.
Now suppose that $p$ is not the characteristic polynomial of $T$. Then it must be a multiple of the minimal polynomial, and because $\deg p=\dim V$ this would imply that there is some factor $(z-\lambda)^r$ in $p$ where $r$ is bigger than the multiplicity of $\lambda$, but this would imply that there are $T|_{V_j}$ and $T|_{V_k}$ that share some generalized eigenvector of $\lambda$. But this is impossible because $T$ is invariant in each $V_k$, what means that if $v\in V_k$ then $(w I-T)^nv\in V_k$ for all $n\in\Bbb N$ and any chosen $w\in\Bbb C$.
Hence $p$ is the characteristic polynomial of $T$.
