Let, $ f(x)=sinh \;x \; \; \therefore f(0)=sinh \; 0 \; = \; 0 \\ \; \\ \; \\ Differentiating \; w.r.t. \;x, \; repeatitively, \\ \; \\ \therefore f_1(x) \; = \; cosh \;x \; \; \therefore f_1(0) \; = \; cosh \; 0 \; = \; 1 \\ \; \\ \; \\ \therefore f_2(x) \; = \; sinh \;x \; \; \therefore f_2(0) \; = \; 0 \\ \; \\ \; \\ \therefore f_3(x) \; = \; cosh \;x \; \; \therefore f_3(0) \; = \; 1 \\ \; \\ \; \\ \therefore f_4(x) \; = \; sinh \;x \; \; \therefore f_4(0) \; = \; 0 \\ \; \\ \; \\ \therefore f_5(x) \; = \; cosh \;x \; \; \therefore f_5(0) \; = \; 1 \\ \; \\ \; \\ \; \\ By \; using \; Maclaurin's \; series, \; \\ \; \\ f(x)\;=\; f(0)+xf_1(0)+\dfrac{x^2}{2!}f_2(0)+ \dfrac{x^3}{3!}f_3(0)+\ldots+\dfrac{x^n}{n!}f_n(0) \\ \; \\ \; \\ \therefore sinh \;x \; = \; 0+x(1)+\dfrac{x^2}{2!}(0)+ \dfrac{x^3}{3!}(1)+ \dfrac{x^4}{4!}(0)+ \dfrac{x^5}{5!}(1)+ \dfrac{x^6}{6!}(0)+ \dfrac{x^7}{7!}(1)+\ldots \\ \; \\ \; \\ \; \\ \therefore sinh \; x \; = \; x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\ldots \; Hence \; Proved. $