Question : Let $a_1, \dots, a_n, e_1, \dots, e_n$ be arbitrary integers. And let $p_1, \dots, p_n$ be distinct primes. Show that there exists $a$ such that $v_p(a - a_i) = e_i$ for $i = 1,\dots, n$.
I would like to know if my reasoning is correct : I have tried using the Chinese Remainder Theorem proving that there exists $a$ such that : $$ a = a_i\text{ mod } p_i^{e_i} $$ That is true since all $p_i^{e_i}$ are relatively prime. Then I would have that: $$ a = q_i.p_i^{e_i} + a_i$$ for some $q_i$. So, $$ v_p(a - a_i) = v_p(p_i^{e_i}) + v_p(q_i) = e_i + v_p(q_i) $$ But here I need to prove that we have $v_p(q_i) = 0$ for $i = 1, \dots, n$. And I don't know if that is true, and how to prove it. Any ideas?
Here is an alternate approach to prove what you're asking for. First, let
$$j = \prod_{i=1}^{n} p_i^{e_i} \tag{1}\label{eq1A}$$
Now, for any $1 \le i \le n$, you can use the Chinese Remainder Theorem to prove there exists an $a$ such that
$$a \equiv a_i + j \pmod{p_i^{e_{i} + 1}} \tag{2}\label{eq2A}$$
As you stated in your approach, this can be done since all $p_i^{e_{i} + 1}$ are relatively prime. Thus, for some $q_i$, you have
$$a = q_i\left(p_i^{e_{i} + 1}\right) + a_i + j = p_i^{e_{i}}\left(q_i(p_i) + \frac{j}{p_i^{e_{i}}}\right) + a_i \tag{3}\label{eq3A}$$
Let
$$k_i = q_i(p_i) + \frac{j}{p_i^{e_{i}}} \tag{4}\label{eq4A}$$
Since $p_i \not\mid \frac{j}{p_i^{e_{i}}}$ (since all $p_i$ are distinct primes), then $p_i \not\mid k_i$. You get
$$v_p(a - a_i) = v_p(p_i^{e_i}) + v_p(k_i) = e_i + 0 = e_i \tag{5}\label{eq5A}$$