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The lecturer has not taught us proofs yet, so I think the question is not looking for a rigorous proof. My attempt:

$$30 = 2 \times 3 \times 5$$ $$\frac{30^n}{14} = \frac{2^n \times 3^n \times 5^n}{2 \times 7} = \frac{2^{n-1} \times 3^n \times 5^n}{7}$$

Now I have written this in prime numbers, which I presume will make it easier to solve.
I feel that $7$ is not going to divide any power of $2$, $3$, or $5$.
This is as close as I can get to a proof.

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  • $\begingroup$ Yeah, it seems alright. Just take mod 7 on both sides and you're done. $\endgroup$ – Matti P. Mar 13 '19 at 11:15
  • $\begingroup$ The proof is absolutely valid. $\endgroup$ – Peter Mar 13 '19 at 11:17
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Your idea is fine, but you don't have to actually divide $30^n$ by $14$. Just note that $30^n=2^n3^n5^n$ (which you did) and that therefore, since there is no $7$ here (and since $7$ is prime), $2^n3^n5^n$ is not a multiple of $7$. In particular, it is not a multiple of $14$.

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  • $\begingroup$ My confusion is: How do we know $7$ cannot divide $2^n3^n$ or $2^n5^n$ $3^n5^n$, these combinations? $\endgroup$ – wingerse Mar 13 '19 at 11:35
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    $\begingroup$ Because if a prime divides a product, it divides one of the factors. So, if $7\mid2^n3^n5^n$, then $7\mid2$, or $7\mid3$, or $7\mid5$. But this does not occur. $\endgroup$ – José Carlos Santos Mar 13 '19 at 11:38
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You can also reason as follows:

  • $30^n = (2\cdot 14 + 2)^n = m\cdot 14 + 2^n$ for a positive integer $m$
  • It follows if $14 | 30^n$, then $14| 2^n$ which is impossible as $7$ is not a factor of any power $2^n$
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We have $$30^n\equiv 2,4,8\mod 14$$ so exists no such exponent for $$n\geq1 $$ and $$n\in\mathbb{N}$$

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