Number Theory Question 2: Proving pn | mn Using Prime Factorization

[trap101]

Let p be a prime and let m and n be natural numbers. Prove that p | mⁿ implies pⁿ | mⁿ.

Attempt:

Since mⁿ can be written out as a product of primes i.e: p₁p₂...p_n in which p is a divisor.

Raising mⁿ means that there would exist pⁿ primes for each factor of m: mⁿ = m₁m₂...m_n = (p₁...p_n)₁(p₁...p_n)₂...(p₁...p_n)_n = p₁^a1p₂^a2...p_n^an

which means pⁿ | mⁿ.

Is there anything I'm missing to clean it up?

Cheers for the help.

 

[fauboca]

It seems ok but could be written better.

Like you said m is a product of primes so let's write m as $m=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_t^{\alpha_t}$. Then $m^n = \left(p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_t^{\alpha_t}\right)^n = p_1^{n\alpha_1}p_2^{n\alpha_2}\ldots p_t^{n\alpha_t}$. Since $p|m^n$, $p|p_i$ for $1\leq i\leq t$.

Spoiler

Let $p_j$ be that p. Then $p|p_j$. Since $p_j$ is prime, $p=p_j$.

$p|m^n=p_1^{n\alpha_1}p_2^{n\alpha_2}\ldots p_j^{n\alpha_j} p_t^{n\alpha_t}$.

$n\alpha_j\geq n$ so $p^n|p_j^{n\alpha_j}\Rightarrow p^n|m^n$

 

[Deveno]

i would prove p|m first. then pⁿ|mⁿ is rather obvious.

in intuitive terms, we don't get "any new prime factors" when we take a power.

again p|mⁿ → p|m is a simple application of the fact:

p|ab → p|a or p|b, which some people take as a definition of prime (it generalizes well to more general settings).

(a "rigorous" proof of p|mⁿ→p|m would use induction on n, but i feel that's over-kill).