Some good lemma a for power of 4 (NT) Prove that for any positive integer $n$ there are $n$ consecutive composite numbers all less than $4^{n+2}$.
I want to use the lemma:    For any prime numbers less than
a positive integer $k$ , product of those primes strictly less than $4^k$.
 A: I don't offhand see any direct way to use your stated lemma that the product of all of the prime numbers less than a positive integer $k$ is less than $4^k$. Instead, here is an alternate way to solve your problem.
The requirement that there are at least $n$ consecutive composite numbers means there are no primes within those $n$ values, i.e., the prime gap must be at least $n + 1$. For there to be no such $n$ consecutive composites means all prime gaps must be at most $n$. Thus, there will be at least one set of $n$ consecutive composites before $4^{n+2}$ if the average prime gap (i.e., the length divided by the # of primes) is greater than $n$. This can be expressed as
$$\frac{4^{n+2}}{\pi\left(4^{n+2}\right)} \gt n \implies \frac{4^{n+2}}{n} > \pi\left(4^{n+2}\right) \tag{1}\label{eq1A}$$
where $\pi(x)$ is the prime-counting function.
The Non-asymptotic bounds section of Wikipedia's "Prime number theorem" page gives [1] that for $x \ge 55$ you have
$$\frac{x}{\log(x) - 4} \gt \pi(x) \tag{2}\label{eq2A}$$
Now, for $x = 4^{n+2}$, having $\log(x) - 4 \gt n$ means the LHS of \eqref{eq2A} is less than the LHS of \eqref{eq1A}, so it proves \eqref{eq1A} holds. This required condition simplifies to
$$\begin{equation}\begin{aligned}
\log\left(4^{n+2}\right) - 4 & \gt n \\
(n+2)\log(4) - 4 & \gt n \\
\log(4)(n) + 2\log(4) - 4 & \gt n \\
(\log(4) - 1)n & \gt 4 - 2\log(4) \\
n & \gt \frac{4 - 2\log(4)}{\log(4) - 1} \approx 3.17
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
For $n = 1$, there are already $5$ consecutive composite integers of $24$ to $28$ occurring before $4^{n+2} = 64$. Also, note that $64 \gt 55$, so \eqref{eq2A} applies. Also, automatically for all $n \le 5$, you have those $5$ consecutive composites occurring before $4^{n+2}$, so your statement holds. For $n \gt 5$, you have \eqref{eq3A} confirming you still have at least $n$ consecutive composites occurring before $4^{n+2}$.
References:
[1] Rosser, Barkley, Explicit bounds for some functions of prime numbers, Am. J. Math. 63, 211-232 (1941). ZBL0024.25004.
