Definitions:
Let $\xi_n$ be a symmetric random walk, i.e., $$ \xi_n=\eta_1+\eta_2+\dots+\eta_n, $$ where $\{\eta_n\}$ is a sequence of i.i.d. random variables such that $$ P\{\eta_n=1\}=P\{\eta_n=-1\}=\frac{1}{2}. $$ Furthermore, we define the first hitting time to be $$\tau=\min\left\{n:|\xi_n|=K\right\},$$ where $K$ is a positive integer.
I was reading a book on stochastic processes and here we want to show that $\tau<\infty$ a.s. The book proves this as follows
We want to show that $P\{\tau=\infty\}=0.$ To this end we shall estimate $P\{\tau>2Kn\}.$ Notice that $$P\{\tau>2Kn\}\le \left(1-\frac{1}{2^{2K}}\right)^n\longrightarrow 0$$ as $n\to\infty.$ Thus, we have \begin{align} P\{\tau=\infty\}&=\bigcap_{n=1}^\infty P\{\tau>2Kn\} \\ &=\lim_{n\to\infty} P\{\tau>2Kn\}=0. \end{align}
After spending some time, I could not figure out how to get the inequality $$P\{\tau>2Kn\}\le \left(1-\frac{1}{2^{2K}}\right)^n$$ in the first line of the proof. Can someone help me understand why this inequality holds?
Many thanks in advance.
$\frac{1}{2^{2K}}$ is the probability of $2K$ successive $+1$s in a row in an attempt of $2K$ steps. If this happens you must either hit the upper boundary or must have started below the lower boundary, since they are $2K$ apart.
$1-\frac{1}{2^{2K}}$ is the probability of not having $2K$ successive $+1$s in a row in an attempt of $2K$ steps.
$\left(1-\frac{1}{2^{2K}}\right)^n$ is the probability of not having $2K$ successive $+1$s in a row in any of $n$ independent attempts of $2K$ steps ($2Kn$ steps in total).
That last probability is therefore less than or equal to the probability of never hitting the boundary in $2Kn$ steps, which is $P\{\tau>2Kn\}$.
Tighter bounds are possible but unnecessary.