Splitting 1.1 and using binomial theorem to write the first few terms we have $(1.1)^{10000}$ = $(1+0.1)^{10000}$ = $\,{}^{10000}C_0+\,{}^{10000}C_1(0.1)$ + $\,{}^{10000}C_2(0.1)^2$ + ... + $\,{}^{10000}C_{10000}(0.1)^{10000}$ = 1 + 10000 × $\frac{1}{10}$ + some positive terms = 1 + 1000 + some positive terms > 1000. Hence, $(1.1)^{10000}$ is larger than 1000.