Both sides are solutions of $\,f(n\!+\!1)-f(n) = (n\!+\!1)(n\!+\!1)!,\ f(0) = 0,\:$ so they are equal by the uniqueness theorem for first-order recurrences, a.k.a. the fundamental theorem of difference calculus. This has a trivial one-line inductive proof (expressed in the language of sums vs. recurrences it boils down to telescopic cancellation of sums of differences).
As I often emphasize, uniqueness theorems provide powerful tools for proving equalities.
On the general topic of functions defined by mathematical induction ("inductive or recursive definition"), see the award-winning Monthly exposition by Leon Henkin, On mathematical induction, AMM, 1960.
Remark $\ $ Note that some of the other answers simply give the standard inductive proof, using ellipses rather than a formal inductive argument. This is most certainly not a "proof without using mathematical induction". These proofs are all standard mechanical applications of telescopy.
A proof that a statement is true for all integers must - at some point or another - employ mathematical induction. The use of induction may not be obvious - it may be hidden (far) down the inference chain in some other theorem or lemma invoked, as in said uniqueness theorem for recurrences (difference equations).
The idea of using uniqueness theorems certainly does differ from the standard inductive proof. Indeed, the uniqueness theorem for higher-order recurrences may be viewed as a generalization of the vivid telescopic cancellation that occurs in the first-order case. One can find some examples in the linked posts.