Induction: $\sqrt{1+\sqrt{n+\sqrt{n^2+\sqrt{n^3+...+\sqrt{n^n}}}}} < n$ Prove using induction for all natural $n>1$.
I can't seem to find a way to use the assumption (for $n$) to prove the $n+1$ case.
Thanks ahead.
 A: Let $$a_0=\sqrt{n^n},\ \ \ a_{k}=\sqrt{n^{n-k}+a_{k-1}},\ \ \ k=1,\ldots,n.$$ The estimate we are looking for is $a_n<n$. 
Put $b_1=\sqrt2$. We have, since $n>2$, that $n-1>n/2$, so  $$ a_1=\sqrt{n^{n-1}+n^{n/2}}\leq\sqrt{2n^{n-1}}=b_1\,n^{(n-1)/2}$$
In general, put $b_{k+1}=\sqrt{1+b_k}$. Assume, as inductive step, that $$a_k\leq b_k\,n^{(n-k)/2}.$$ Then, if $n\geq k+2$, we have $n-k-1\geq(n-k)/2$, and so 
\begin{align}
a_{k+1}&=\sqrt{n^{n-k-1}+a_k}\leq\sqrt{n^{n-k-1}+b_k\,n^{(n-k)/2}}
\leq\sqrt{n^{n-k-1}+b_k\,n^{n-k-1}}\\ \ \\ &=\sqrt{1+b_k}\,n^{\frac{n-k-1}2}
= b_{k+1}\,n^{\frac{n-(k+1)}2}.
\end{align}
Because of the condition $n\geq k+2$, we can continue this process until $k=n-2$, i.e. 
$$
a_{n-1}\leq b_{n-1}\,n^{\frac{n-(n-1)}2}=b_{n-1}\,\sqrt n.
$$
Then
$$\tag{1}
a_n=\sqrt{1+a_{n-1}}\leq\sqrt{1+b_{n-1}\,\sqrt n}
\leq b_n\,\sqrt{1+\sqrt n}
$$
Note that the sequence $\{b_k\}$ is increasing. Also, $b_1<2$ and, if $b_k<2$, then $$b_{k+1}=\sqrt{1+b_k}<\sqrt{1+2}=\sqrt3<2.$$ So the sequence is also bounded, and is thus convergent. It is easy to check that the limit is $\varphi=\frac{1+\sqrt5}2$, and then we have $$b_k<\varphi=\frac{1+\sqrt5}2,\ \ k=1,2,\ldots $$
Then $(1)$ becomes 
$$
a_n<\varphi\,\sqrt{1+\sqrt n}.
$$
Now, for $n\geq3$, 
$$
\varphi\,\sqrt{1+\sqrt2}<n
$$
(it's easy to check the plot),
and so we get $a_n<n$, as desired. The case $n=2$ can be checked manually, namely
$$
\sqrt{1+\sqrt{2+\sqrt{2^2}}}=\sqrt{1+2}=\sqrt3<2.
$$
