To prove,

1(1!) + 2(2!) + 3(3!) + ... + n(n!) = (n+1)! - 1

LHS = 1(1!) + 2(2!) + 3(3!) + ... + n(n!)

RHS = (n+1)! - 1

By mathematical induction:

**Let n = 1**,

Then, LHS = 1(1!) = 1 x 1 = 1

And RHS = (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1

So, both LHS and RHS = 1 and equation is true at n=1.

Now, second part is the assumption part where n=k holds true for the equation.

**Let n = k**,

The equation is assumed to be...

(The entire section contains 163 words.)

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