All of the coefficients of $P$ are at most $5$. If one of the coefficients is $5$, then $P(x)=5x^n$, which is impossible since $177$ is not a power of $5$. Therefore, all of the coefficients are less than $5$, so since $177=1202_5$, we must have $P(x)=x^3+2x^2+2$, so $P(10)=\boxed{1202}$.

We have $P(5)=177$. Let $n$ be the degree of $P$. If $n\geq 4$ then $P(5)\geq a_n 5^n\geq 5^4>177$, a contradiction. So, $n\in \{0,1,2,3\}$. It's clear that $n$ can't be $0$ or $1$. If $n=2$ then let $P(x)=a_2x^2+a_1x+a_0$. Now, $$a_2+a_1+a_0=5, \ 25a_2+5a_1+a_0=177.$$By mod $5$ we get $a_0=2$ and after solving we get $a_1=-5$, this is absurd. If $n=3$ then let $P(x)=a_3x^3+a_2x^2+a_1x+a_0$. So, $$a_3+a_2+a_1+a_0=5, \ 125a_3+25a_2+5a_1+a_0=177.$$We get $a_0=2$ also the second equation says $a_3\leq 1$ so $a_3=1$. Now solving we get $a_2=2, a_1=0$. Thus $P(x)=x^3+2x^2+2$. So $P(10)=10^3+2\cdot 10^2+2=\boxed{1202}$.

Let p(x) = a_0 + a_1 x + a_2 x² + ......... + a_n x^n where a_0,a_1,a_2,.........,a_n are nonnegative integers. p(1) = 5 and p(5) = 177 a_0 + a_1 + a_2 + ........ + a_n = 5 a_0 + 5a_1 + 25a_2 + 125a_3 + ..........+ 5^n.a_n = 177 a_0 = 2(mod 5) a_0 = 2 since p(1) = 5. It's easy to observe that a_4 = a_5 = .......... = a_n = 0 since 5⁴ > 125. 5a_1 + 25a_2 + 125a_3 = 175 a_1 + 5a_2 + 25a_3 = 35 a_1 = 0(mod 5) a_1 = 0 5a_2 + 25a_3 = 35 a_2 + 5a_3 = 7 a_2 = 2(mod 5) a_2 = 2 and a_3 = 1 Therefore, p(x) = x³ + 2x² + 2 p(10) = 1000 + 200 + 2 = 1202