Integration using Incomplete Gamma function

Question

Evaluate ${I_1}\left( z \right) = \int\limits_0^z {{{\left( {1 - x} \right)}^{b - 1}}{e^{ - ax}}} dx;\,a,b > 0\,,0 < z < 1$
I have solved using incomplete Gamma function.
By a substitution $\left( {x - 1} \right)a = t$, the above integration becomes
$= \frac{1}{a}\int\limits_{ - a}^{a(z - 1)} {{{\left( { - \frac{t}{a}} \right)}^{b - 1}}{e^{ - a\left( {\frac{t}{a} + 1} \right)}}} dt$\ $= \frac{{{e^{ - a}}}}{{{{\left( { - a} \right)}^b}}}\int\limits_{ - a}^{a(z - 1)} {{t^{b - 1}}{e^{ - t}}} dt$
By definition of incomplete Gamma function $\gamma \left( {a,x} \right) = \int\limits_0^x {{e^{ - t}}} {t^{a - 1}}dx$
=$ \frac{{{e^{ - a}}}}{{{{\left( { - a} \right)}^b}}}\left[ {\gamma \left( {b,a(z - 1)} \right) - \gamma \left( {b, - a} \right)} \right]$. Any errors in my method of solving.