The measure of the image of a linear map

Question

A somewhat similar question was already asked but I'm trying to prove it a different way. I'm somewhat new to the website so not sure if I should've commented on the old thread or not but here it is. I'm working through measure theory and I'm stuck on the following exercise.

Suppose $T\in\text{GL}_d(\mathbb{R})$. The goal is to prove that for any borelian $B\in\mathcal{B}(\mathbb{R})$, we have $\lambda_d(T(B)) = |\det(T)|\lambda_d(B)$, where $\lambda_d$ is the $d$-dimensional Lebesgue measure. The exercise has four steps:

1. Show there exists $k_T>0$ such that for any borelian $B\in\mathcal{B}(\mathbb{R})$, we have $\lambda_d(T(B))=k_T\lambda_d(B)$.
2. Verify that for $S,T\in\text{GL}_d(\mathbb{R})$, $k_{ST}=k_Sk_T$.
3. Show the result above when $T$ is diagonal, then when $T$ is symmetric and finally, when $T$ is orthogonal.
4. Conclude using the fact that for $T\in\text{GL}_d(\mathbb{R})$, there exists $U$ orthogonal and $S$ symmetric such that $T=US$.

I honestly don't know where to start. Even for the first step I'm kind of lost. Apparently, from linearity I can show this for the unit cube and then it naturally extends to all borelian but I don't understand why this would be true either. Any help would be appreciated!