Answer
$$\pi $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^\pi } - 1}}{h} \cr
& {\text{The limit can be written as }} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^\pi } - {{\left( {1 + 0} \right)}^\pi }}}{{h - 0}} \cr
& {\text{From the definition of the derivative }} \cr
& f'\left( c \right) = \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}} \cr
& \underbrace {\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^\pi } - {{\left( {1 + 0} \right)}^\pi }}}{{h - 0}}}_{f'\left( c \right) = \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}}} \Rightarrow f\left( h \right) = {\left( {1 + h} \right)^\pi },{\text{ }}a = 0 \cr
& {\text{Then }} \cr
& f'\left( h \right) = \frac{d}{{dh}}\left[ {{{\left( {1 + h} \right)}^\pi }} \right] \cr
& f'\left( h \right) = \pi {\left( {1 + h} \right)^{\pi - 1}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^\pi } - {{\left( {1 + 0} \right)}^\pi }}}{{h - 0}} = f'\left( 0 \right) \cr
& f'\left( 0 \right) = \pi {\left( {1 + 0} \right)^{\pi - 1}} = \pi {\left( 1 \right)^{\pi - 1}} = \pi \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^\pi } - 1}}{h} = \pi \cr} $$
