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synkk

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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