evaluating $\Gamma(1/2)$ using elementary methods Can we prove $\Gamma(1/2)$=$\sqrt{\pi}$ using elementary techniques .
I can easily prove properties of gamma functions like $\Gamma(n+1)$=n$\Gamma(n)$and value of $\Gamma(1)$ using integration by parts .
Most of the  proves i saw used the integral $\int_0^\infty e^{-x^2}$ , wheres to solve this integral i substitute $x^2$=t and then get the answer as (1/2)$\Gamma(1/2)$=$\sqrt{\pi}/2$ becuase i am a high school student and know only elementary methods to solve integrals
Non conventional methods /some other beautiful methods (which can be elementary or near to elementary) to evaluate this  are appreciated .
 A: Recall the Beta function
$$\int_0^{\pi/2}\cos^{2x-1}(\theta)\sin ^{2y-1}(\theta)d \theta= \frac{1}{2}\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \tag{1}$$
now, let $x=\frac{1}{2}\,\, \text{and}\,\,y=\frac{1}{2}$ in $(1)$
$$
\begin{aligned}
\int_0^{\pi/2}d \theta= \frac{1}{2}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}\\
\frac{\pi}{2}=\frac{1}{2}\Gamma^2\left(\frac{1}{2}\right)\\
\Gamma^2\left(\frac{1}{2}\right)=\pi\\
\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}
\end{aligned}
$$
A: Over $(-1,1)$ we have
$$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} x^n \tag{1}$$
and since the $\Gamma$ function allows to extend factorials to $\mathbb{R}^+$, it also allows to extend binomial coefficients, such that
$$ \frac{1}{4^n}\binom{2n}{n} = \binom{-1/2}{n}(-1)^n\tag{2} $$
must hold for any $n\in\mathbb{N}$ by the binomial theorem, providing the Maclaurin series of $(1+x)^\alpha$. In terms of the $\Gamma$ function we are stating
$$ \frac{1}{4^n}\frac{\Gamma(2n+1)}{\Gamma(n+1)^2}=(-1)^n \frac{\Gamma(1/2)}{\Gamma(n+1)\Gamma(1/2-n)}, \tag{3}$$
but
$$ \frac{1}{\Gamma(1/2-n)} = (-1)^n \frac{\Gamma(1/2+n)}{\pi}\tag{4} $$
due to the reflection formula, so
$$ \frac{1}{4^n}\frac{\Gamma(2n+1)}{\Gamma(n+1)}=\frac{\Gamma(1/2)\Gamma(1/2+n)}{\pi }\tag{5} $$
and by considering the instance $n=0$ we have $\Gamma(1/2)^2=\pi$, so $\Gamma(1/2)=\sqrt{\pi}$ since $\Gamma$ is positive over $\mathbb{R}^+$.

Alternative approach: the area of the unit circle is $\pi$ by definition, so
$$ \pi=4\int_{0}^{1}\sqrt{1-x^2}\,dx \stackrel{x\mapsto\sqrt{z}}{=} 2 \int_{0}^{1} z^{-1/2}(1-z)^{1/2}\,dz. \tag{6}$$
The RHS of $(6)$ is twice the value of a Beta function, precisely
$$ 2\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} = \Gamma(1/2)^2\tag{7} $$
and the conclusion is the same as before.
A: \begin{align}\Gamma\left(\frac{1}{2}\right)&=\int_0^\infty x^{-\frac{1}{2}}\text{e}^{-x}dx\\
&\overset{y=\sqrt{x}}=2\int_0^\infty \text{e}^{-y^2}dy\\
\left(\frac{\Gamma\left(\frac{1}{2}\right)}{2}\right)^2&=\left(\int_0^\infty \text{e}^{-y^2}dy\right)\left(\int_0^\infty \text{e}^{-t^2}dt\right)\\
&=\int_0^\infty \int_0^\infty\text{e}^{-(t^2+y^2)}dtdy\\
&\overset{t(u)=uy}=\int_0^\infty \int_0^\infty y\text{e}^{-y^2(u^2+1)}dudy\\
&=\int_0^\infty \left[-\frac{\text{e}^{-y^2(u^2+1)}}{2(1+u^2)}\right]_0^\infty du\\
&=\frac{1}{2}\int_0^\infty \frac{1}{1+u^2}du\\
&=\frac{\pi}{4}\\
\Gamma\left(\frac{1}{2}\right)&=\boxed{\sqrt{\pi}}
\end{align}
