For n=7 we have
3^7<7!<-->2187<5040 (true)
We shall prove that if the
statement is true for n=k
(k=integer) then it is also
true for the next integer,
in other words for n=k+1
So suppose that
3^k<k! is true (1)
Then we must prove that
3^(k+1)<(k+1)! (2)
Multiplying both sides of (1)
with 3 we have
3^(k+1)<3k!
So in order the (2) statement
be true must happens
3k!<(k+1)!<-->
3K!<K!(K+1)<-->
3<K+1<-->
K>2 which is obviously
true
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For n=7 we have
3^7<7!<-->2187<5040 (true)
We shall prove that if the
statement is true for n=k
(k=integer) then it is also
true for the next integer,
in other words for n=k+1
So suppose that
3^k<k! is true (1)
Then we must prove that
3^(k+1)<(k+1)! (2)
Multiplying both sides of (1)
with 3 we have
3^(k+1)<3k!
So in order the (2) statement
be true must happens
3k!<(k+1)!<-->
3K!<K!(K+1)<-->
3<K+1<-->
K>2 which is obviously
true