Reasoning about the congruence classes modulo $p_n\#$ related to $(\frac{p_n\#}{p_i})^k$ Let $p_n$ be the $n$th prime and $p_n\#$ the primorial for $p_n$.
Is it correct to assume that for any $i \le n$ and $w$, there exists $a < p_i$ such that:
$$\left(\frac{p_n\#}{p_i}\right)^w \equiv a\left(\frac{p_n\#}{p_i}\right) \pmod {p_n\#}$$
If this is true, how does one prove it?  If it is not true, could you provide a counter example?

Edit:  I think that I found an argument that it is true in all cases.
Apologies for the number of steps, I am open to any suggestions to simplify it or correct it.
Lemma: if $x \equiv b \pmod {ab}$, then for any $w$, $\exists{c}$ such that $c < a$ and $x^w \equiv cb \pmod {ab}$
(1)  $\exists y$ such that $x = aby + b$
(2)  $x^2 = a^2b^2y^2 + 2ab^2y + b^2 = ab(aby^2 + 2by) + b^2$
(3)  If $b > a$, $\exists m,n$ such that $b = am+n$ with $n < a$
(4)  $b^2 = b(am+n) = abm + nb$ so that $x^2 \equiv nb \pmod {ab}$
(5)  Assume that this is true up to $w$ so that $x^w \equiv nb \pmod {ab}$
(6)  $x^{w+1} \equiv x(nb) \equiv nb^2 \equiv n^2b \pmod {ab}$
(7)  If $n^2 > a$, $\exists s,t$ such that $n^2 = as+t$ with $t < a$
(8)  $n^2b = abs + tb$
(9)  So that $x^{w+1} \equiv tb \pmod {ab}$ with $t < a$
Corollary: $\left(\frac{p_n\#}{p_i}\right)^w \equiv a\left(\frac{p_n\#}{p_i}\right) \pmod {p_n\#}$ with $a < p_i$
(1)  Let $x = \frac{p_n\#}{p_i}$ so that $b = \frac{p_n\#}{p_i}$ and $a = p_i$
(2)  The conclusion follows.
 A: Your conjecture and lemma are both basically correct, but there are a number of issues with your proof of the lemma, and in one case particular case the applicability of it to your conjecture based on your proof.
First, with your conjecture, if $w = 0$, then it's not true except with $n = i = 1$. It seems you're implicitly assuming $w \ge 1$. Also, you didn't put a lower bound on $a$, but I believe you want it to be $0 \le a \lt p_i$ (although the $0$ case never applies, so perhaps $1$ was your intended lower bound instead).
With your lemma statement, the same issues with $w$, and $c$ instead of $a$, as I discuss in my above paragraph, also apply to it. In the proof of your lemma,

(3)  If $b > a$, $\exists m,n$ such that $b = am+n$ with $n < a$

You don't actually need the restriction $b \gt a$ since if $b \lt a$, then $m = 0$ with $n = b$, and if $b = a$, then $m = 1$ with $n = 0$. Also, you should include the implicit $0 \le n \lt a$ since it seems you're using the Euclidean division lemma here.

(5)  Assume that this is true up to $w$ so that $x^w \equiv nb \pmod {ab}$

The $n$ used here will not in general be the same as in your $(3)$. For example, $x = 5$, $b = 5$ and $a = 3$ gives $ab = 15$ and $b = 3(1) + 2$, so $n = 2$. However, $w = 3$ gives $5^3 = 125 \equiv 5 \equiv (1)5 \pmod{15}$, so the $n$ here would be $1$ instead of $2$. I suggest using a different variable in your $(5)$, say $q$, with a statement such as $0 \le q \lt a$.
Also, it seems you are trying to use a form of induction proof when you wrote to assume a statement is true for up to $w$ and then proving it for $w + 1$. However, in your lemma statement, although $w$ is a non-specified positive integer, it's fixed for any particular use of the lemma. In your $(5)$ statement, though, it appears to be something which is varying since, later, you handle $x^{w+1}$ but the lemma only deals with $x^{w}$. This confusion can be avoided by using a different later variable, with $k$ being a common one in induction proofs, instead of $w$.

(7)  If $n^2 > a$, $\exists s,t$ such that $n^2 = as+t$ with $t < a$

This step doesn't explicitly state what is to happen when $n^2 \le a$, with $s$ is then technically being undefined, although it's used in $(8)$, and similarly $t$ not being defined but is used in $(8)$ and $(9)$.
Finally, with your corollary,

(1)  Let $x = \frac{p_n\#}{p_i}$ so that $b = \frac{p_n\#}{p_i}$ and $a = p_i$

If $n = i = 2$, then $b = \frac{6}{3} = 2$ and $a = 3$, so your condition of $b \gt a$ in $(3)$ means the lemma technically doesn't apply.
Here's how I would prove your lemma instead. With $x \equiv b \pmod {ab}$, then $ab \mid x - b \implies b \mid x - b \implies b \mid x$. Thus, there's a $k \in \mathbb{Z}$ where $x = kb$. Then there exists $m, c \in \mathbb{Z}$ where $k^{w}b^{w-1} = ma + c$ with $0 \le c \lt a$. This gives
$$x^{w} = (kb)^{w} = (k^{w}b^{w-1})b = (ma + c)b = mab + cb \equiv cb \pmod{ab} \tag{1}\label{eq1A}$$
This lemma can then be used to prove your conjecture for all cases. However, there's another method by using a lemma which is more general in terms of the congruence values, but it's more restricted in terms of the factors of the modulus, although it supports more than $2$ factors being used. This is

Lemma #$1$: Let the $r \ge 2$ non-zero integers $n_i$ for $1 \le i \le r$ be such that $\gcd(n_i, n_j) = 1 \; \forall \; 1 \le i, j \le r, i \ne j$. Then for integers $x$ and $y$
$$x \equiv y \pmod{\prod_{i=1}^{r}n_i} \iff x \equiv y \pmod{n_i} \; \forall \; 1 \le i \le r \tag{2}\label{eq2A}$$
Proof: This was basically asked and answered in Show that $a\equiv b \pmod{n_1n_2...n_r}$ if and only if $a \equiv b\pmod {n_i}$ for all $i = 1, 2,. . . , r$. I believe the answer there could be made a bit clearer using that for any integer $q$ we have
$$x \equiv y \pmod{q} \iff x - y \equiv 0 \pmod{q} \iff q \mid x - y \tag{3}\label{eq3A}$$
Then for $\implies$ in \eqref{eq2A}, since $(\prod_{i=1}^{r}n_i) \mid x - y$, then $n_i \mid x - y \; \forall \; 1 \le i \le n$. With $\impliedby$, we get for some $k_i \in \mathbb{N}, 1 \le i \le r$ that $x - y = k_{1}n_{1} = k_{2}n_{2} = \ldots = k_{r}n_{r}$. This means for all $2 \le i \le r$ that $n_i \mid k_{1}n_{1}$. Since $\gcd(n_i, n_1) = 1$, this means $n_i \mid k_1$, so combining this for all $i$ gives $(\prod_{i=2}^{r}n_i) \mid k_1 \implies n_1(\prod_{i=2}^{r}n_i) \mid k_{1}n_{1}$, so $(\prod_{i=1}^{r}n_i) \mid x - y$.

With $r = 2$, $n_1 = p_i$ and $n_2 = \frac{p_n\#}{p_i}$, your conjecture then becomes to show, for all integers $w \ge 1$, there's an integer $0 \le a \lt n_1$ such that
$$n_{2}^{w} \equiv an_{2} \pmod{n_{1}n_{2}} \tag{4}\label{eq4A}$$
Using my lemma #$1$ gives the equivalent requirements of
$$n_{2}^{w} \equiv an_{2} \pmod{n_{1}} \tag{5}\label{eq5A}$$
$$n_{2}^{w} \equiv an_{2} \pmod{n_{2}} \tag{6}\label{eq6A}$$
Similar to my proof for your lemma, there exists $m, a \in \mathbb{Z}$ where $n_{2}^{w - 1} = mn_{1} + a$ with $0 \le a \lt n_{1}$. Then
$$n_{2}^{w} = (n_{2}^{w - 1})n_{2} = (mn_{1} + a)n_{2} = mn_{1}n_{2} + an_{2} \equiv an_{2} \pmod{n_{1}} \tag{7}\label{eq7A}$$
Thus, \eqref{eq5A} is true, with the value of $a$ also working in \eqref{eq6A} since it gives $0 \equiv a(0) \pmod{n_{2}}$. Thus, \eqref{eq4A} is also true, showing your conjecture to be true.
