calculus - Limit with Gamma function

Let $\Gamma (s)=\int_0^\infty t^{s-1}e^{-t} dt$ - classical gamma function - and $\gamma(s,x)=\int_0^x t^{s-1}e^{-t} dt$ - lower incomplete gamma function. Then is the following true

$$\lim_{\alpha \to 1} \frac{1}{\Gamma(1-\alpha)} \gamma\left( 1-\alpha, [\Gamma(1-\alpha)]^{-1/\alpha} \right)=1$$
or false?

Intuitively, looking at the expression of $\gamma$ and taking into account that $\lim_{\alpha\to1}\Gamma(1-\alpha)=\infty$, I would say the limit is $0$. Yet, if we can use the fact that $\gamma(s,x)$ near $x=0$ behaves like $x^s/s$, this would lead us to

$$\begin{equation} \begin{split} &\lim_{\alpha \to 1} \frac{1}{\Gamma(1-\alpha)} \gamma\left( 1-\alpha, [\Gamma(1-\alpha)]^{-1/\alpha} \right)\\ & \quad=\lim_{\alpha \to 1} \frac{[\Gamma(1-\alpha)]^{-1/\alpha}}{1-\alpha} \\ & \quad=\lim_{\alpha \to 1} \frac{[\Gamma(\alpha)\sin(\pi\alpha)/\pi]^{1/\alpha}}{1-\alpha}\\ & \quad=\lim_{\alpha \to 1} \frac{[\sin(\pi\alpha)/\pi]^{1/\alpha}}{1-\alpha}. \end{split} \end{equation}$$
Then, using the fact that when $x$ is close to $\pi$, $\sin(x) \approx \pi -x$, I would go on with
$$\begin{equation} \hspace{-4em}=\lim_{\alpha \to 1} \frac{[(\pi -\pi \alpha)/\pi]^{1/\alpha}}{1-\alpha}\\ \hspace{-6.4em}=\lim_{\alpha \to 1} \frac{(1-\alpha)^{1/\alpha}}{1-\alpha}\\ \hspace{.7em}=\lim_{\alpha \to 1} \exp \left\lbrace \left(\frac{1}{\alpha}-1\right)\log(1-\alpha) \right\rbrace \end{equation}$$
which eventually equals $1$, since the exponent in the last line converges to $0$. Am I loosing something/making some mistake?

Answer

Hint:

$$\frac1{x[\Gamma(x)]^{1/(1-x)}}=\frac{x^{x/(1-x)}}{[x\Gamma(x)]^{1/(1-x)}}=\frac{x^{x/(1-x)}}{[\Gamma(x+1)]^{1/(1-x)}}\to_0\frac1{[\Gamma(1)]^1}=1$$

Since

$$\lim_{x\to0}x^{x/(1-x)}=1$$