Let $2=p_1<p_2<\cdots$ be the sequence of primes in increasingorder.Prove that

\begin{equation}

p_n\leq 2^{2^{n-1}}

\end{equation}for all $n\geq 1$.

Proof:When $n=1$,$2\leq 2$.Suppose $\forall n\leq k$,

\begin{equation}

p_n\leq 2^{2^{n-1}}

\end{equation}

Then

\begin{equation}

p_{n+1}\leq p_1p_2\cdots

p_n+1\leq 2^{1+2^1+\cdots+2^{n-1}}+1(p_1p_2\cdots p_n+1~\mbox{is a prime.})=2^{2^n-1}+1\leq 2^{2^n}

\end{equation}

By induction,$\forall n\in\mathbf{N}^{+}$,

\begin{equation}

p_n\leq 2^{2^{n-1}}

\end{equation}