Below is my effort to provide an alternative argument for the Sylvester-Schur theorem that is simpler than that of Paul Erdős.

Please let me know if I made mistakes or if a point is unclear.

The argument uses the following lemma:

**lemma**: For integers $ n ge $ 9 and $ x ge 2n $::

$$ {x select n}> dfrac {x!} {(x – pi (n))!} $$

(1) For $ n ge 8 $It's easy to show that:

$$ n> ( ln n) ^ 2 $$

Note: You can find details here $ n> ( ln n) ^ 2 $

(2) For $ n ge $ 9, it follows $ (2 ln n)> 4.25506 $ and since $ n> ln n $::

$$ 2n ln (n) – left ( dfrac {n + 2} {n + 1} right) ln n – 1.25506n> 4.25506n – left ( dfrac {n + 2} {n + 1} right) ln n – 1.25506n> n $$

(3) $ 2n ln n – left ( dfrac {n + 2} {n + 1} right) ln n – 1.25506n> ( ln n) ^ 2 $

(4) $ (2n-1) – dfrac {1.25506n} { ln n}> ln n + dfrac {1} {n + 1} $

(5) Multiply $ n + 1 $ on both sides

$$ (n + 1) left ((2n-1) – dfrac {1.25506n} { ln n} right)> (n + 1) ln n + 1 $$

(6) Multiply $ ln n $ on both sides:

$$ (n + 1) ((n-1) ln n – 1.25506n)> ( ln n) ((n + 1) ln n -n + 1) $$

(7) $ n + 1> left ( dfrac { ln n} {(n-1) ln n – 1.25506n} right) left ((n + 1) ln n – (n-1) ln e right) $

(8th) $ sqrt (n – 1 – frac {1.25506n} { ln n}) { left (2n – dfrac {1.25506n} { ln n} right) left (2n-1 – dfrac { 1.25506n} { ln n} right) points (n + 1)}> (n + 1)> sqrt (n – 1 – frac {1.25506n} { ln n}) { dfrac {n ^ {n + 1}} {e ^ {n-1}}} $

(9) Since $ n! le dfrac {n ^ {n + 1}} {e ^ {n-1}} $, it follows:

$$ sqrt (n – 1 – frac {1.25506n} { ln n}) { left (2n – dfrac {1.25506n} { ln n} right) left (2n-1 – dfrac {1.25506n} { ln n} right) points (n + 1)}> (n + 1)> sqrt (n – 1 – frac {1.25506n} { ln n}) {n!} $$

Note: See sentence here for the argument that $ n! le dfrac {n ^ {n + 1}} {e ^ {n-1}} $.

(10) Since $ x ge 2n $::

$$ left (x – dfrac {1.25506n} { ln n} right) left (x-1 – dfrac {1.25506n} { ln n} right) dots (xn + 1) ge left (2n – dfrac {1.25506n} { ln n} right) left (2n-1 – dfrac {1.25506n} { ln n} right) dots (n + 1)> n ! $$

(11) As for $ n> 1 $, $ pi (n) < dfrac {1.25506n} { ln n}: $

$$ (x- pi (n)) (x-1- pi (n)) points (n + 1)> n! $$

Note: See 3.6, Sentence 2, Corollary 1, by Rosser & Schoenfeld, 1962

(12) $ dfrac {(x) (x-1) points (x-n + 1)} {n!}> (x) (x-1) points (x- pi (n) +1) $

(13) It follows:

$$ {x choose n}> frac {x!} {(x – pi (n))!} $$

**sentence**: For integers $ n ge 9, x ge 2n $there is a prime number $ p> n $ so that $ p | dfrac {x!} {(x-n)!} $

(1) Leave $ x, n $ be integers so that $ n ge $ 9 and $ x ge 2n $.

(2) Suppose there is no prime number greater than $ n $ that divides $ dfrac {x!} {(x-n)!} $

(3) Leave $ text {lcm} (x-n + 1, x-n + 2, dots, x) $ be their least common multiple.

(4) It can be shown that:

$$ text {lcm} (x-n + 1, x-n + 2, dots, x) ge {x choose n} $$

Note: The argument for this can be found here

(5) From the assumption in (2) it follows:

$$ text {lcm} (x-n + 1, x-n + 2, dots, x) < dfrac {x!} {(x – pi (n))!} $$

Note: where $ pi (n) $ is the prime count function.

(6) But then we have a contradiction since e.g. $ n ge 9, x ge 2n $follows from the lemma:

$$ {x select n}> dfrac {x!} {(x – pi (n))!} $$

Feel free to let me know if a point is unclear or if there is a better way to make the same argument.