MHB Prove $10^{\dfrac{5^n -1}{4}} \Huge\vert (5^n)!$ for $Z^+$

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The discussion revolves around proving that \(10^{\frac{5^n - 1}{4}} \) divides \((5^n)!\) for all positive integers \(n\). Participants clarify that the vertical bar indicates divisibility, not a fraction. Legendre's formula is introduced as a potential tool for the proof, which calculates the exponent of a prime in a factorial. The conversation suggests starting with \(n=1\) as a base case and then applying mathematical induction to establish the general case. The consensus is that the proof should be straightforward once the initial case is confirmed.
karush
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prove
$10^{\dfrac{5^n -1}{4}} \Huge\vert (5^n)!$
for all $Z^+$

ok this was sent to me on email but thot I could solve it but ?

first I assume the vertical bar means such that

not sure if ! means factorial or not negative

anyway curious
 
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$a \mid b$ means $a$ divides $b$.

The following formula may help.

Legendre's formula: For any prime number $p$ and any positive integer $m$, let
${\displaystyle \nu _{p}(m)}$ be the exponent of the largest power of $p$ that divides $m$. Then

$${\displaystyle \nu _{p}(m!)=\sum _{i=1}^{\infty }\left\lfloor {\frac {m}{p^{i}}}\right\rfloor}$$
 
ok i have never tried that
are you sure the vertical bar means a/b?
 
karush said:
ok i have never tried that
are you sure the vertical bar means a/b?
Not a/b it's a divides into b evenly. For example, 3|15 and 5|15.

-Dan
 
oh
can we prove this just by setting n=1
 
karush said:
oh
can we prove this just by setting n=1
I don't know what you mean.

It's true for $n=1$ because $10$ divides $120$.
 
karush said:
oh
can we prove this just by setting n=1
From there use Mathematical Induction. It should be rather easy but I haven't really spent much time thinking about it. (ie. If it works for n = 1 then you always have that factor.)

-Dan
 

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