Answer
$f\left( x \right) = {{\rm{e}}^{x - 1}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( {x + 1} \right)}^n}}}{{{{\rm{e}}^2}n!}}$
Work Step by Step
Write ${{\rm{e}}^{x - 1}} = {{\rm{e}}^{\left( {x + 1} \right) - 2}} = \dfrac{1}{{{{\rm{e}}^2}}}{{\rm{e}}^{\left( {x + 1} \right)}}$.
We have from Table 2 of Section 11.7:
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting $\left( {x + 1} \right)$ for $x$ in ${{\rm{e}}^x}$, we get
${{\rm{e}}^{\left( {x + 1} \right)}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( {x + 1} \right)}^n}}}{{n!}}$, ${\ \ \ }$ converges for all $x$.
Thus, the Taylor series centered at $c=-1$ is
$f\left( x \right) = {{\rm{e}}^{x - 1}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( {x + 1} \right)}^n}}}{{{{\rm{e}}^2}n!}}$.
It is valid for all $x$.
