Recall the Binomial Theorem formula.

The binomial expansion of $(a+b)^n$ is given by:

$$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

In this case, $a=1$, $b=\frac{1}{x}$, and $n=5$.

The terms are $\binom{5}{k} (1)^{5-k} (\frac{1}{x})^k$ for $k=0, 1, 2, 3, 4, 5$.

For $k=0$:

$$\binom{5}{0} (1)^{5} \left(\frac{1}{x}\right)^0 = 1 \cdot 1 \cdot 1 = 1$$

For $k=1$:

$$\binom{5}{1} (1)^{4} \left(\frac{1}{x}\right)^1 = 5 \cdot 1 \cdot \frac{1}{x} = \frac{5}{x}$$

For $k=2$:

$$\binom{5}{2} (1)^{3} \left(\frac{1}{x}\right)^2 = 10 \cdot 1 \cdot \frac{1}{x^2} = \frac{10}{x^2}$$

For $k=3$:

$$\binom{5}{3} (1)^{2} \left(\frac{1}{x}\right)^3 = 10 \cdot 1 \cdot \frac{1}{x^3} = \frac{10}{x^3}$$

For $k=4$:

$$\binom{5}{4} (1)^{1} \left(\frac{1}{x}\right)^4 = 5 \cdot 1 \cdot \frac{1}{x^4} = \frac{5}{x^4}$$

For $k=5$:

$$\binom{5}{5} (1)^{0} \left(\frac{1}{x}\right)^5 = 1 \cdot 1 \cdot \frac{1}{x^5} = \frac{1}{x^5}$$

$$(1 + \frac{1}{x})^5 = 1 + \frac{5}{x} + \frac{10}{x^2} + \frac{10}{x^3} + \frac{5}{x^4} + \frac{1}{x^5}$$

The simplified binomial expansion is:

$$\boxed{1 + \frac{5}{x} + \frac{10}{x^2} + \frac{10}{x^3} + \frac{5}{x^4} + \frac{1}{x^5}}$$

$$(0.05)^5 = (5 \times 10^{-2})^5$$ $$(0.05)^5 = 5^5 \times (10^{-2})^5$$ $$(0.05)^5 = 3125 \times 10^{-10}$$ $$(0.05)^5 = 0.0000000003125$$

To round to 4 decimal places, we look at the fifth decimal place.
The number is 0.0000000003125.
The first four decimal places are $0, 0, 0, 0$.
The fifth decimal place is 0.
Since the fifth decimal place is 0 (which is less than 5), we keep the fourth decimal place as it is.

$$0.0000$$

The value of $(0.05)^5$ correct to 4 decimal places is:

$$\boxed{0.0000}$$