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## Homework Statement

Prove the [tex]\lim_{n\to +\infty}{\sqrt[n]{ n! }} \equiv \infty [/tex]

## Homework Equations

uses well-known operations

## The Attempt at a Solution

I think the best (easiest) approach is to find some [tex] f(n) \leq n! [/tex] , and express it as some [tex]g(n)^{n}[/tex] . This will then get rid of the annoying n-root, and it should then be easy to show that [tex]\lim_{n\to +\infty}{g(n)} = \infty [/tex] , which implies the limit is infinity for [tex]n![/tex] (i.e. the squeeze theorem only with regard to the lower bound).

Since [tex]n! = (n)(n-1)(n-2) ... (n-n+2) [/tex] , I thought of using the smallest multiple [tex]( n - n + 2)^{n-1}[/tex], but I still cannot express this as a function to the n power, and even if I did, the limit would just be 2.. so it's smaller than [tex]n![/tex], but not 'big enough'.

I think I may need a different approach. Suggestions?