User:The-problem

$$\frac{\tan\left[45-\frac{1}{2}\tan^{-1}\left(\frac{2n-1}{2n^2-2n}\right)\right]} {\tan\left[\frac{1}{2}\tan^{-1}\left(\frac{2n-1}{2n^2-2n}\right)\right]}=\frac{(n-1)(2n-1)}{n}$$

odd
$$1\cdot{3}\cdot{5}\cdot{2^3}=5!$$

$$5\cdot{7}\cdot{9}\cdot{2^4}=7!$$

2
$$\frac{3!^2+4!^2}{1!+2!+3!+4!+5!}=2^2$$

$$3!^3+4!^3=\frac{5!^2}{1!^2}-\frac{6!^1}{2!^1}$$

$$1^2\left(1!^2+2!^2+3!^2+4!^2+5!^2\right)+2^2\left(1!+2!+3!+4!+5!\right)=5^{3!}+2^{2!}$$

$$\frac{4!^3+5!^3}{4!^2+5!^2}=\frac{2!\left(6!+6^2\right)}{1!+2!\cdot{3!}}$$

Cube
$$2^3-2=3!\cdots(1)$$

$$3^3-3=4!\cdots(2)$$

$$5^3-5=5!\cdots(3)$$

$$9^3-9=6!\cdots(4)$$

64
$$6^4=6!+(4!)^2\cdots(1)$$

333
$$4\times4!-4-3\times3!-3=3!^2+3!^2+3\cdots(1)$$

Square
$$1^2+2^2+3^2+4^2+5^2+4^2+3^2+2^2+1^2=6^2+7^2\cdots(1)$$

Cube
$$3\times3!-3-2\times2!-2=1^3+2^3\cdots(1)$$

$$5\times5!-5-4\times4!-4=3^3+5^3+7^3\cdots(3)$$

Factorial
$$n!+n!+1=X^2\cdots(1)$$

$$4!+4!+1=7^2\cdots(2)$$

$$n!+n!=X^2-X\cdots(1)$$

$$1!+1!=2^2-2\cdots(2)$$

$$3!+3!=4^2-4\cdots(3)$$

$$5!+5!=16^2-16\cdots(4)$$

Cube
$$2^Y-Y=X^3\cdots(0)$$

$$2^1-1=1^3\cdots(1)$$

$$2^5-5=3^3\cdots(2)$$

Factorial
$$2^X-{2^Y}=N!\cdots(0)$$

$$2^1-{2^0}=1!\cdots(1)$$

$$2^2-{2^1}=2!\cdots(2)$$

$$2^3-{2^1}=3!\cdots(3)$$

$$2^5-2^3=4!\cdots(4)$$

$$2^7-2^3=5!\cdots(5)$$