Which is bigger: $(\pi+1)^{\pi+1}$ or $\pi^{\pi+2}$? I've been struggling for a while with the following problem:
Which is bigger: $(\pi+1)^{\pi+1}$ or $\pi^{\pi+2}$?
Needless to say software aid is not allowed. All manual calculations should be possible to be performed in reasonable time (for example, I did read some solutions to similar problems that to my understanding include calculating 113th power of a decimal number, and this does not count as "reasonable" manual calculation).

Probably useless for the solution, but nevertheless interesting graphs of  $(x+1)^{x+1}-x^{x+2}$:




Using computer:
$(\pi+1)^{\pi+1} = 359.796$
$\pi^{\pi+2} = 359.867$
 A: Unfortunately, needles are needed. This work is for $(\pi + 1)/\pi$, not $(\pi+2)/(\pi + 1)$, if you are going to make it more precise, third order approximation is needed.
I think needles are not necessary. $a = \pi$, We try to prove 
$\ln(a + 1)/\ln(a) < (\frac{a+1}{a})$.
well, $a \sim 355/113\sim 3$, the error on estimate of $\pi$ is very tiny, will not harm our result, we do $a = 3 + b$, thus $b = 16/113\sim 1/7 \sim 0.14$
we just show it for
$\frac{\ln(4) + \ln(1+ b/4)}{\ln(3) + \ln(1 + b/3)}$
we use Taylor expansion on $\ln(1 + b/4)$ and $\ln(1 + b/3)$ to the first order term. 
The following is estimate of error from truncation.
$$\vert\frac{u}{v} - \frac{u + s}{v + t}\vert \le \frac{|ut| + |vs|}{v(v+t)}$$
since $\frac{u}{v}$ is around $4/3 < 2$, $v$ is bounded below by $1$, we know that the error will be bounded by 
$2(|t| + |s|) \le 2(b/3)^2 \le 1/100$.
$\ln(1 + b/4) \sim b/4 \sim 0.0354$
$\ln(1 + b/3) \sim b/3 \sim 0.0472$
$\ln(4) = 2\ln(2) \sim 1.386$
$\ln(3) \sim 1.099$
result is $\frac{1.386 + 0.0354}{1.099 + 0.0472} = 1.2400 < (\pi + 1)/(\pi) = 1.32$
By the way, 
$\log(\pi + 1)/(\log(\pi)) \sim 1.2414$
A: Consider the function f(x) =($\frac{x+1}{x}$)^(x+1);  the sign of the derivative on [3, 4] depends of the expression xln($\frac{x+1}{x})$ - 1 and both f ‘(3) and f ‘(4) are negative and not null on the interval. Hence f is decreasing over [3, 4]. We have f(3) = 3.160493827 > $\pi$  and f(4) = 3.051757813 < $\pi$ . We note that f(3) is nearer of $\pi$ than f(4); calculate f(3.14) = ($\frac{414}{314})^{4.14}$ = 3.141178 < $\pi$. Therefore f($\pi$) < $\pi$. Consequently the second expression is bigger.             
A: Let $x=(\pi+1)^{\pi+1}$ and $y=\pi^{\pi+2}$
Since $\ln x=(\pi+1)ln(\pi+1)$ and $\ln y=(\pi+2)\ln\pi$ and $\ln x$ is increasing ,
compare $\ln x$ and $\ln y$
So I depend on a internet calculating, $\ln x-\ln y<0$ and so $x<y$
But $\ln x-\ln y=-0.00019...$ tell us that it is difficult to clear for manually.
(I tried to prove for " $f(x)=(x+1)\ln(x+1)-(x+2)\ln x$ is decreasing and $f(\pi)<0$" 
But $f(3.14)>0$ ,according to calculating)
