What is the proof for a^n > n for all natural numbers n and a >= 2?

[synkk]

Prove for all $x \geq -1$ and n being a natural number, $(1+x)^n \geq 1 + nx$ I've done this using induction

hence deduce from this that for any real number # a \ geq 2 $,$ a^n > n ## for all n belonging to the natural numbers.

I'm stuck at the deduce part

I'm not sure what deduce means, it seems fairly obvious so instead I tried to prove it, by again, induction:

a = 2,
2^n > n which is true a n >= 1
assume true for a = k

then when a = k+1

(k+1)^n > n

I don't see how I can prove it from here, it seems obvious as k >= 2 and n >=1 this will always hold true, and thus a^n > n but I can't seem to prove it. Does anyone have any tips on proving as I'm fairly new at this

thank you

 

[pasmith]

Prove for all $x \geq -1$ and n being a natural number, $(1+x)^n \geq 1 + nx$ I've done this using induction

hence deduce from this that for any real number # a \ geq 2 $,$ a^n > n ## for all n belonging to the natural numbers.

I'm stuck at the deduce part

I'm not sure what deduce means

It means "show that this follows from what you have just proved".

In this case, you've shown that if x \geq -1 then (1 + x)^n \geq 1 + nx.

Now see what happens if you set a = 1 + x.

 

[synkk]

It means "show that this follows from what you have just proved".

In this case, you've shown that if x \geq -1 then (1 + x)^n \geq 1 + nx.

Now see what happens if you set a = 1 + x.

how would that work exactly? it states that a >= 2 if we let a = 1 + x then a >= 0 as x >= -1?

 

[synkk]

if I ignore my confusion above I ge tthis:

a^n >= 1 + nx

as x >= -1 a^n >= 1 - n

as n is a natural number n >=1
hence a^n > n