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## Main Question or Discussion Point

I understand that the standard proof is a bit different from my own, but I want to know if my reasoning is valid. PROOF:

Firstly, I assume that x is positive.

I then consider p = inf{n∈ℕ : n>x} . In other words, I choose "p" to be the smallest natural number greater than x. If we choose n>p, then we can rewrite the original statement as ((x/1)(x/2)...(x/(p-1)))*((x/p)...(x/n)). If we let α=((x/1)(x/2)...(x/(p-1))), then x

Is this subsequent method for finding sufficiently large "n" viable in practice, and are my findings valid?

Firstly, I assume that x is positive.

I then consider p = inf{n∈ℕ : n>x} . In other words, I choose "p" to be the smallest natural number greater than x. If we choose n>p, then we can rewrite the original statement as ((x/1)(x/2)...(x/(p-1)))*((x/p)...(x/n)). If we let α=((x/1)(x/2)...(x/(p-1))), then x

^{n}/n! = α((x/p)...(x/n)) where each term x/k is less than one. Thus α((x/p)...(x/n)) < α(x/n). So if we want to require that (x^{n}/n!) < ε for some ε>0, then we can choose n satisfying α(x/n) < ε ⇒ n > (αx/ε). Since α = (x^{p-1}/(p-1)!), we must have n > (x^{p}/ε(p-1)!). Thus, any n satisfying n > max[p, (x^{p}/ε(p-1)!)] will ensure that x^{n}/n! < ε.Is this subsequent method for finding sufficiently large "n" viable in practice, and are my findings valid?