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Let $phi_1, ldots phi_d: mathbb{R} to mathbb{R}$ be continuos and have compact support.
Show that for
$$
f := bigotimes_{i = 1}^{d} phi_i: mathbb{R}^d to mathbb{R},
(x_1, ldots, x_d) mapsto prod_{i = 1}^{d} phi_i(x_i)
$$
we have
$$
int_{mathbb{R}^d} f(x) dx = prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
$$

Proof
Since $f$ is continuous and has a compact support as well, we get
begin{align*}
int_{mathbb{R}^d} f(x) dx
& = int_{text{supp}(f)} prod_{i = 1}^{d} phi_i(x_i) dx
overset{(ast)}{=} int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 1}^{d} phi_i(x_i) dx_1 right) ldots right) d x_d \
& overset{(star)}{=} left( int_{text{supp}(phi_1)} phi_1(x_1) dx_1 right) left( int_{text{supp}(phi_d)} left( ldots left( int_{text{supp}(phi_1)} prod_{i = 2}^{d} phi_i(x_i) dx_2 right) ldots right) d x_d right) \
& = ldots
= prod_{i = 1}^{d} int_{mathbb{R}} phi_i(x_i) dx_i
end{align*}

My Question
I don't understand the steps $(ast)$ and especially $(star)$ . Can we just ''pull out'' a one-dimensional integral since it's just a constant?

The proof is from Otto Forster: Analyis 3 (in German) on page 3.
This was a homework assignment and the corrector gave full marks to this answer but suggested using induction instead of the $ldots$-step.
Could someone please show that induction in light of the question above?

Please bear in mind that this is from the very beginning of real analysis III and we have not learned measure theory yet.

Yes, you can pull out $intphi_1(x_1), dx_1$ because it is a constant. To use induction simply note that when you pull out $intphi_1(x_1), dx_1$ you are left with a similar integral with $n-1$ variables instead of $n$. By induction hypothesis this integral w.r.t $n-1$ variables can be written as a product of $n-1$ integrals.