# prove that 1(1^1)+2(2^2)+...n(n!)=(n+1)!-1 for all positive integers

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## 1(1^1)+2(2^2)+...n(n!)=(n+1)!-1

L.H.S= 1(1^1)+2(2^2)+...+n(n!)

n=1......L.H.S=1

n=1......R.H.S=(1+1)!-1

=2!-1

=2-1

=1

So when n=1 this is true

n=2...L.H.S=1(1^1)+2(2^2)

=5

n=2...R.H.S=(2+1)!-1

=3!-1

=6-1

=5

So when n=2 this is true

we take this is true when n=p (**p** is a real number)

so **1(1^1)+2(2^2)+...+p(p!)=(p+1)!-1** is true

n=(p+1)...L.H.S=**1(1^1)+2(2^2)+...+p(p!)+**(p+1)(p+1)!

put **1(1^1)+2(2^2)+...+p(p!)=(p+1)!-1**

=(p+1)!-1**+**(p+1)(p+1)!

=(p+1)!**+**(p+1)(p+1)!-1

=(p+1)!{1+(p+1)}-1

=(p+1)!(p+2)-1

=(p+2)!-1

so acording the math logical 1(1^1)+2(2^2)+...+n(n!)=(p+1)!-1 is true