Show that if $2^t \le (t+1)^n, n\ge 5$, then $ t \le n^2-1$

Question

I am reading a book in which the following is just stated without proof: If $2^t \le (t+1)^n, n\ge 5$, then$ t \le n^2-1$. I've tried expanding the RHS using the binomial theorem, but I haven't gotten anywhere with it. I'm especially puzzled by the condition that $n\ge5$. A crude way would be to plug in $n=5$, which I am soon to attempt, but I would like to know if there is another way. Any hints in the right direction?

Answer 1

HINT: $t\ln 2\leq n\cdot\ln(t+1)$.

Answer 2

Assume $t>n^2-1$. Then $t\ge n^{1.97}$ and from $$t\ln 2\le n\ln(t+1) $$ we get $$n^{1.97}\ln 2< t\ln 2\le n\ln(n^2) =2n\ln n$$ so $$\tag1n^{0.97}\le \frac 2{\ln 2}\cdot\ln n<3\ln n.$$ But the power on the left grows faster than the log on the right, making $(1)$ false for $n\ge 5$.

We conclude that $t\le n^2-1$ if $n\ge 5$.