How to prove $(1-\frac1{36})^{25}\lt\frac12$? How to prove the inequality? 
$(1-\frac1{36})^{25}\lt\frac12$

I'm in trouble. 
Thank you very much for your help

Here I present a basic proof only using the binomial expansion theorem without using the property of $(1-\frac{1}{x})^x$. Note that the inequality is equivalent to the following inequality
$$ \left(\frac{36}{35}\right)^{25}> 2.$$
By the binomial expansion theorem, we have
\begin{eqnarray}
\left(\frac{36}{35}\right)^{25}&=&\left(1+\frac{1}{35}\right)^{25}\\
&>&1+\binom{25}{1}\frac{1}{35}+\binom{25}{2}\frac{1}{35^2}+\binom{25}{3}\frac{1}{35}\\
&=&1+\frac{25}{35}+\frac{25\cdot 24}{2}\frac{1}{35^2}+\frac{25\cdot24\cdot23}{3\cdot 2}\frac{1}{35^3}\\
&=&1+\frac{5}{7}+\frac{12}{49}+\frac{92}{1715}\\
&=&1+\frac{1737}{1715}\\
&>&2.
\end{eqnarray}
Done.
