Find an asymptotic bound for recurrence $T(n)=T(n-1)+n(n-1)$

Question

Find an asymptotic bound for recurrence $T(n)=T(n-1)+n(n-1)$.

I tried 2 methods and I'm getting very different results.

Method 1: $$ T(n)=n(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+4\cdot 3+3\cdot 2+2\cdot 1=\\ =\sum_{i=0}^n (n-i)(n-i-1)=\sum_{i=0}^n n^2-n(2i+1)+i^2+i=\ast $$ The dominant term here is $n^2$ so: $$ \sum_{i=0}^n n^2=\frac{2n^3+3n^2+n}{6}\le n^3 $$ Therefore $T(n)=T(n-1)+n(n-1)\le n^3$.

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Method 2: $$ \frac{T(n)}{n!}=\frac{n(n-1)}{n!}+\frac{T(n-1)}{n!}=\\ =\frac{1}{(n-2)!}+\frac{1}{(n-3)!}+...+\frac{1}{3!}+\frac{1}{2!}+1=\\ =\sum_{i=1}^n \frac{1}{i!}\le e-1\Rightarrow T(n)\le n!(e-1) $$

Worlfram Alpha yet gives an answer which is different from the two above.

Answer 1

Wolfram says that the solution for the recurrence relation is $t(n)=c_1+\frac{1}{3}n(n^2-1)$, so method 1 works. For method 2,

1) Why do you have $-\frac{T(n-1)}{n!}$ after first equality?

2) If your goal is to find the tightest bound, the solution for method 1 is better. Method 2 also works but the resulting bound isn't the tightest one.