What is the easiest way to see that the following statement is true?
Suppose $Y_1, \dots, Y_n \overset{\text{iid}}{\sim} \text{Exp}(1)$. Show $\sum_{i=1}^{n}(Y_i - Y_{(1)}) \sim \text{Gamma}(n-1, 1)$.
Note that $Y_{(1)} = \min\limits_{1 \leq i \leq n}Y_i$.
By $X \sim \text{Exp}(\beta)$, this means that $f_{X}(x) = \dfrac{1}{\beta}e^{-x/\beta} \cdot \mathbf{1}_{\{x > 0\}}$.
It is easy to see that $Y_{(1)} \sim \text{Exponential}(1/n)$. Furthermore, we also have that $\sum_{i=1}^{n}Y_i \sim \text{Gamma}(\alpha = n, \beta = 1)$ under the parametrization $$f_{Y}(y) =\dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}\mathbf{1}_{\{x > 0\}}\text{, }\qquad \alpha, \beta> 0\text{.}$$
Solution given Xi'an's Answer: Using the notation in the original question: $$\begin{align} \sum_{i=1}^{n}[Y_i - Y_{(1)}] &= \sum_{i=1}^{n}[Y_{(i)}-Y_{(1)}] \\ &= \sum_{i=1}^{n}Y_{(i)}-nY_{(1)}\\ &= \sum_{i=1}^{n}\{Y_{(i)}-Y_{(i-1)}+Y_{(i-1)}-\cdots-Y_{(1)}+Y_{(1)}\}-nY_{(1)}\\ &= \sum_{i=1}^n\sum_{j=1}^{i}\{Y_{(j)}-Y_{(j-1)}\}-nY_{(1)}\text{ where } Y_{(0)} = 0 \\ &= \sum_{j=1}^n\sum_{i=j}^{n}\{Y_{(j)}-Y_{(j-1)}\}-nY_{(1)}\\ &= \sum_{j=1}^{n}(n-j+1)[Y_{(j)}-Y_{(j-1)}]-nY_{(1)}\\ &= \sum_{i=1}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]-nY_{(1)}\\ &= \sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]+nY_{(1)}-nY_{(1)} \\ &= \sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]\text{.} \end{align}$$ From this, we get that $\sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}] \sim \text{Gamma}(n-1, 1)$.