Since ${\bf P}^n$ is the quotient of ${\bf A}^{n+1} - 0$ by the action of ${\bf G}_m$, the tangent bundle of ${\bf P}^n$ is the quotient of the tangent bundle of ${\bf A}^{n+1} - 0$ by the action of the tangent bundle of ${\bf G}_m$:
$T {\bf P}^n = T ({\bf A}^{n+1} - 0) / T {\bf G}_m$ .
As a group, the tangent bundle of ${\bf G}_m$ is the product of ${\bf G}_m$ and a 1-dimensional vector space V. Therefore we can take the quotient of everything on the right side above by ${\bf G}_m$. Note that $T({\bf A}^{n+1}-0)$ is the product of ${\bf A}^{n+1} - 0$ with the direct sum of (n+1) copies of the weight one representation of ${\bf G}_m$. Therefore its quotient by ${\bf G}_m$ is $\mathcal{O}_{{\bf P}^n}(1)^{n+1}$. We get
$T {\bf P}^n = \mathcal{O}_{{\bf P}^n}(1)^{n+1} / V$ .
A (linear) action of a one-dimensional vector space on a vector bundle is the same thing as an exact sequence, so this gives the Euler sequence.
