Cut elimination is indispensable for studying fragments of arithmetic. Consider for example the classical Parsons–Mints–Takeuti theorem:

**Theorem** If $I\Sigma_1\vdash\forall x\,\exists y\,\phi(x,y)$ with $\phi\in\Sigma^0_1$, then there exists a primitive recursive function $f$ such that $\mathrm{PRA}\vdash\forall x\,\phi(x,f(x))$.

The proof goes roughly as follows. We formulate $\Sigma^0_1$-induction as a sequent rule
$$\frac{\Gamma,\phi(x)\longrightarrow\phi(x+1),\Delta}{\Gamma,\phi(0)\longrightarrow\phi(t),\Delta},$$
include axioms of Q as extra initial sequents, and apply cut elimination to a proof of the sequent $\longrightarrow\exists y\,\phi(x,y)$ so that the only remaining cut formulas appear as principal formulas in the induction rule or in some axiom of Q. Since other rules have the subformula property, *all* formulas in the proof are now $\Sigma^0_1$, and we can prove by induction on the length of the derivation that existential quantifiers in the succedent are (provably in PRA) witnessed by a primitive recursive function given witnesses to existential quantifiers in the antecedent.

Now, why did we need to eliminate cuts here? Because even if the sequent $\phi\longrightarrow\psi$ consists of formulas of low complexity (here: $\Sigma^0_1$), we could have derived it by a cut
$$\frac{\phi\longrightarrow\chi\qquad\chi\longrightarrow\psi}{\phi\longrightarrow\psi}$$
where $\chi$ is an arbitrarily complex formula, and then the witnessing argument above breaks. Sigfpe wrote above in his answer that cut elimination makes proofs more complex, but that’s not actually true: cut elimination makes proofs *longer*, but more *elementary*, it eliminates complex concepts (formulas) from the proof. The latter is often useful, and it is the primary reason why so much time and energy is devoted to cut elimination in proof theory. In most applications of cut elimination one does not really care about having no cuts in the proof, but about having control of which formulas can appear in the proof.