The random vector $(\xi_1,\dots, \xi_n)^T$ is uniformly distributed in $$D = \{0 \leq x_1 \leq \dots \leq x_n \leq 1\}$$
I need to find the distribution of $(\xi_1, \xi_2−\xi_1,\dots, \xi_n−\xi_{n-1})^T$
I know, that if $\xi$ has a distribution density $p_\xi (x_1,\dots, x_n)$, then we can consider the joint distribution of $\eta_1 = \xi_1, \eta_j = \xi_i - \xi_{i-1}$ where $i=2,\dots, n$
Then inverse mapping $x_i=y_1+...+y_i$ and $$p_\eta(y_1, \dots, y_n) = p_\xi(y_1, y_1 + y_2, \dots, y_1 + \dots + y_n)$$
But how to find this density?
The volume of region $D$ is $1/n!$, since it equals the probability that $n$ IID uniform $[0,1]$ variables are in sorted order, and there are $n!$ equally likely ways to arrange these variables.
So the joint density of $(\xi_1,\ldots,\xi_n)$, being uniform over $D$, is constant over the region $D$: $$p_\xi(x_1,\ldots,x_n)=n! I_D(x_1,\ldots,x_n)=n! I(0\le x_1\le x_2\le\cdots\le x_n\le1),\tag1$$ where the indicator function $I_A(x)$ has value $1$ when $x\in A$ and $0$ otherwise. You've already computed the joint density for $\eta$ as $$p_\eta(y_1, \dots, y_n) = p_\xi(y_1, y_1 + y_2, \ldots, y_1 + \dots + y_n),$$ so now it's a matter of plugging into (1): $$p_\eta(y_1, \dots, y_n)=n!I(0\le y_1\le y_1+y_2\le\cdots\le y_1+\cdots y_n\le1)\tag2$$ You can rewrite the condition on the RHS of (2) in the equivalent form: $$ 0\le y_1+\cdots+y_n\le 1\qquad\text{and}\qquad 0\le y_i\le 1 \quad\text{for all $i$}.$$