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\begin{document}
\title{5.8 - \#5}
\author{Andrew Lounsbury}
\date{April 1, 2020}
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\section{5.8, p.147}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}
    \item[5.]
    If $a$ is a natural number greater than or equal to 1, then $p_a \le 2^{a^a}$, where $p_a$ is the $a$-th prime number. 
    \begin{proof}
        \underline{Base case}: If $a = 1$, then, $p_a = p_1 = 2 \le 2 = 2^{1^1} = 2^{a^a}$. \\
        \underline{Induction}: Suppose $p_k \le 2^{k^k}$ for some arbitrary natural number $k$. Note that $p_{k+1} \le p_{\min}$, where $p_{\min}$ is the smallest prime factor of $p_1 \cdots p_k-1$. Also, note that $p_1 \cdots p_k = 2p_2 \cdots p_k$ is even, so $p_1 \cdots p_k-1$ is odd. Hence, $p_{k+1} < 3 \le p_{\min}$. We know $p_{k+1} < 3$ because we would otherwise have $3 \le p_{k+1} < p_{\min}$, contradicting $p_{k+1} \le p_{\min}$. Then, $p_{k+1} < 3 < 2^{2^2} \le 2^{(k+1)^{(k+1)}}$, 
        as desired. 
    \end{proof}
\end{enumerate}
\end{document}