The function is given by:

$$f(x) = 1! \cdot x^{2! \cdot x^{3! \cdot x^{4! \cdot x^{5! \cdot x^{6! \cdot x}}}}}$$

We are asked to find the closest value of $f\!\left(\tfrac{1}{2}\right)$.

Solution:

Substitute $x = \tfrac{1}{2}$: $$f\!\left(\tfrac{1}{2}\right) = 1! \cdot \left(\tfrac{1}{2}\right)^{2! \cdot \left(\tfrac{1}{2}\right)^{3! \cdot \left(\tfrac{1}{2}\right)^{4! \cdot \cdots}}}$$

Notice that factorial exponents grow very quickly, so higher terms like $\left(\tfrac{1}{2}\right)^{24}, \left(\tfrac{1}{2}\right)^{120}, \dots$ become extremely small. Thus the tower converges rapidly, and we only need to approximate with the first few terms.

Keeping the dominant contribution: $$f\!\left(\tfrac{1}{2}\right) \approx 1! \cdot \left(\tfrac{1}{2}\right)^{2!} = \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}.$$

Answer: (C) $\;\;\tfrac{1}{4}$