I put this formula on WolframAlpha $$\frac{(26!)^{n+2}}{13!}$$ and it simplified to $$2^{23n+36}\cdot175^{3n+5}\cdot7429^{n+2}\cdot34749^{2n+3}$$

I tried solving it by hand \begin{align} \frac{(26!)^{n+2}}{13!} & = \frac{(26\cdot25\cdot\ldots\cdot3\cdot2)^{n+2}}{13\cdot12\cdot\ldots\cdot3\cdot2} \\\\ & = (26\cdot25\cdot\ldots\cdot15\cdot14)^{n+2} \cdot (13\cdot12\cdot\ldots\cdot3\cdot2)^{n+1} \\\\ & = (2^{13}\cdot3^5\cdot5^4\cdot7^2\cdot11\cdot13\cdot17\cdot19\cdot23)^{n+2}\cdot(2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13)^{n+1} \\\\ & = 2^{23n+36}\cdot(3^5)^{2n+3}\cdot(5^2)^{3n+5}\cdot7^{3n+5}\cdot11^{2n+3}\cdot13^{2n+3}\cdot17^{n+2}\cdot19^{n+2}\cdot23^{n+2} \\\\ & = 2^{23n+36}\cdot(5^2\cdot7)^{3n+5}\cdot(17\cdot19\cdot23)^{n+2}\cdot(3^5\cdot11\cdot13)^{2n+3}\\\\ & = 2^{23n+36}\cdot175^{3n+5}\cdot7429^{n+2}\cdot34749^{2n+3}\\ \end{align}

My question is do I have to do all this work (prime factorization of factorials) everytime I want to simplify an expression of this kind or is there a faster way using Discrete Mathematics?

  • 4
    $\begingroup$ What do you mean by “simplify” in this context? The starting expression seems simpler to my taste. $\endgroup$
    – user208649
    Aug 15, 2020 at 6:25
  • $\begingroup$ How simple do you want the answer? $\endgroup$
    – sai-kartik
    Aug 15, 2020 at 6:27
  • $\begingroup$ Product of exponentials without factorials. I just noticed this is subjective because I can change the way I rearrange the values, so why does WolfremAlpha alwayss solves like this? $\endgroup$ Aug 15, 2020 at 6:27
  • $\begingroup$ hint:the exponent of prime p in $n!$ is given by $[\frac{n}{p}]+[\frac{n}{p^2}]+....$ where[.]is the greatest integer function $\endgroup$ Aug 15, 2020 at 6:36
  • $\begingroup$ Note what Wolfram Alpha did: each prime divides only one of the factors. It grouped the primes into factors by exponent: if the exponents of two primes are constant multiples of the same polynomial in $n$, then those primes are combined into a single factor. $\endgroup$ Aug 15, 2020 at 17:48

1 Answer 1


For more systematic approach, let $v_p(x)$ be a $p$-adic valuation of $x$, then by some basic rules such as $v_p(x^n)=n\cdot v_p(x)$ and $v_p(x/y)=v_p(x)-v_p(y)$. Also we can use Legendre's formula $v_p(n!)=\sum_{i=1}^{\infty}\lfloor \frac{n}{p^i} \rfloor$, so \begin{align} v_2\left(\frac{(26!)^{n+2}}{13!}\right)&=(n+2)v_2(26!)-v_2(13!)\\ &=(n+2)\left(\left\lfloor \frac{26}{2} \right\rfloor + \left\lfloor \frac{26}{4} \right\rfloor + \left\lfloor \frac{26}{8} \right\rfloor + \left\lfloor \frac{26}{16} \right\rfloor \right)\\ &\ \ \ -\left(\left\lfloor \frac{13}{2} \right\rfloor + \left\lfloor \frac{13}{4} \right\rfloor + \left\lfloor \frac{13}{8} \right\rfloor \right)\\ &=(n+2)(13+6+3+1)-(6+3+1)\\ &=23n+36.\\ \end{align} Now just continue in similar manner for $p=3,5,7,11,13,17,19,23$ to obtain complete factorization. Then you can group the terms together in any way you like.


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