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The first step is to express the distribution of the smallest order statistic \(Y_1\). This is because the behavior of \(Y_1\) is of direct interest since it is involved in the definition of the sequence of random variables \(\{Z_n\}\). The distribution function of the smallest order statistic is given by: \(F_{Y_{1}}(y) = 1 - (1 - F(y))^n\). Since \(F\) is the cdf of a continuous random variable \(Y\), in the interval (0,1), defined by the pdf \(f(y)=5y^4\), it follows that: \(F(y) = \int_0^y 5t^4 dt = y^5\). Thus, the distribution function in the case of \(Y_1\) becomes: \(F_{Y_{1}}(y) = 1 - (1 - y^5)^n\). Finally, differentiating \(F_{Y_1}\), the probability density function of \(Y_1\) is given by \(f_{Y_{1}}(y) = 5n y^{4} (1 - y^5)^{n-1}\).

Next, we will define the sequence of random variables of interest and find its limiting distribution. The sequence to study convergence of is \(Z_n = n^p Y_1\). The cumulative distribution function (CDF) of \(Z_n\) can be found by \(P(Z_n <= z) = P(Y_1 <= z/n^p) = F_{Y1}(z/n^p)\) where \(F_{Y1}\) is the CDF of Y_1 as found in step 1. We would like to find a limiting value for the CDF of \(Z_n\) as \(n\) goes to infinity. This is \(lim_{n->\infty} F_{Y1}(z/n^p)\). The function would be \(F_{Y_{1}}(y) = 1 - (1 - \left(\frac{z}{n^p}\right)^5)^n\).

The limiting distribution exists and is non-degenerate when \(lim_{n->\infty} (1- (\frac{z}{n^p})^5)^n\) converges to \(e^{-z^5}\). Using basic analysis of limits, one can recognize that this limiting behavior is similar to \(lim_{n->\infty} (1 + \frac{a}{n})^n\), which equals to \(e^a\) when \(n\) approaches to infinity. Comparing this specific behavior and the process used in the previous step, it can be seen that \(a\), which refers to the rate of exponential decay, equals to \(-z^5\) and \(\frac{-z^5}{n} = \frac{-z}{(n^p)^5}\). As a result, to have similar exponential decay, the denominator should equal \(n\) which results in \(p=1/5\).