One can indeed reduce the inductive proof to verifying the equality $\rm\: n+1 = n+1\:$ by using a special form of induction tailored specifically for such indefinite sums. Namely we can invoke the Fundamental Theorem of Difference Calculus (which has a simple inductive proof).
$\rm\ F(n)\ =\ \sum_{i\:=\:1}^n\ f(i)\ \iff\ \ F(n\!+\!1) - F(n)\ =\ f(n\!+\!1),\ \ \ F(1) =\: f(1)$
In your example we have $\rm\:F(n)\, =\, n(n\!+\!1)/2,\ \ f(n)\, =\, n,\:$ hence $\rm\: F(1) = 1 = f(1),\:$ and
$\rm\ F(n\!+\!1)-F(n)\ =\ \dfrac{(n\!+\!1)(n\!+\!2)}2-\dfrac{n(n\!+\!1)}2\ =\ (n\!+\!1)\left(\dfrac{n\!+\!2}2-\dfrac{n}2\right)\, =\, n\!+\!1\, =\, f(n\!+\!1)$
hence the sought equality follows by the Fundamental Theorem.
Note that by employing the Fundamental Theorem we have reduced the proof to the mechanical verification of the equations on the RHS of the $\iff$, which here amount to checking that two polynomials in $\rm\,n\,$ are equal (here $\rm = n\!+\!1).\:$ No ingenuity is required in devising the inductive step. Instead that ingenuity has been encapsulated once and for all in the proof of the Fundamental Theorem - whose inductive proof is obvious because we have abstracted away from all the peculiarities of its special cases. Namely, the inductive step amounts simply to telescopic cancellation - a general method behind may inductive proofs. For further discussion and examples see my many posts on telescopy.