Series/Sequences help. *question below*

Question

itiaax · Answer

Can I have a walkthrough on this, please? I'm sorta confused by the ! sign in the question

xapproachesinfinity · Answer

okay so what did you do so far?

itiaax · Answer

I've set up u1=1 amd un+1= (n+1)un 
I really dunno how to continue from here as I've never done ! in mathematical induction :x

xapproachesinfinity · Answer

Basically you haven't done anything?

xapproachesinfinity · Answer

start with n=1
we have $\large \rm u_2=2.1=2!$
the case for n=1 is true

xapproachesinfinity · Answer

now we suppose it valid for n
meaning $\large \rm u_n=n!$ is true
then we need to prove the case for n+1 is true
meaning we need to show that $\large \rm u_{n+1}=(n+1)!$ is also true

xapproachesinfinity · Answer

ok can you continue from here

itiaax · Answer

Yes, I suppose I can continue from here. Give me some time :)

xapproachesinfinity · Answer

ok good^_^
any further help tag me

itiaax · Answer

Alrighty :)

xapproachesinfinity · Answer

so far?

xapproachesinfinity · Answer

hint the answer is right in front of you hehe

itiaax · Answer

lol i'm almost finished :)

xapproachesinfinity · Answer

sorry i have to go now! i will check you answer later on^_^

xapproachesinfinity · Answer

i haven't seen your work, but i'm gonna post the solution so you can look at it later

xapproachesinfinity · Answer

so we have $\large \rm u_n=n!~for ~any~n\geq1$ (we supposed it is true
so we have $\large \rm (n+1)u_n=(n+1)n! \Longrightarrow u_{n+1}=(n+1)!$
so it is true for n+1 case as well
and our proof is done
$\square$

itiaax · Answer

Hmm, I had got up to the point where you had (n+1)n! , but I was confused as to how to move on from there to get (n+1), which I knew is what I should end up with. Thank you so much, now I see :) @xapproachesinfinity