Show that there is no natural number $n$ such that $3^7$ is the largest power of $3$ dividing $n!$

Question

Show that there is no natural number $n$ such that $7$ is the largest power $a$ of $3$ for which $3^a$ divides $n!$

After doing some research, I could not understand how to start or what to do to demonstrate this.

We have $$E_3(n!)\neq7\;\;\forall n\in\mathbb{N}\\\left[\frac{n}{3} \right]+\left[\frac{n}{3^2} \right]+\left[\frac{n}{3^3} \right]+\dots\neq7$$I do not know where from, or what to do to solve it.

Answer 1

For $n=15,\left[\frac{15}{3} \right]+\left[\frac{15}{3^2} \right]+\left[\frac{15}{3^3} \right]+\;...=6$
for $n=18$ (the next multiple of $3$) $\left[\frac{18}{3} \right]+\left[\frac{18}{3^2} \right]+\left[\frac{18}{3^3} \right]+\;...=8$
If $n\geq 18$ then $\left[\frac{n}{3} \right]+\left[\frac{n}{3^2} \right]+\left[\frac{n}{3^3} \right]+\;...\geq 8$
So there is no possibility for $7$

Answer 2

Hint: What is the smallest value $n_1$ such that $3^7\mid (n_1)!$? What is the largest value $n_0$ such that $3^7\nmid (n_0)!$? What is the largest exponent $k$ such that $3^k\mid (n_1)!$?