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?